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Every involution of $G{L}_3(2)$ is a transvection $I+vf$, because its nilpotent part has square zero and rank one. Two distinct commuting transvections either have common image line or common fixed plane. This gives two types of elementary abelian groups of order four, displayed above. No larger elementary abelian $2$-subgroup is possible: a transvection commuting with all three elements of a common-image group must have that same image, and the dual argument applies to a common-fixed-plane group. Their normalizers are respectively point and plane stabilizers of order $24$. The common fixed-space dimensions one and two distinguish their conjugacy classes.

For distinct coordinates $i,j$, let $E_{ij}(t)=I+t{e}_{ij}$, where $e_{ij}$ is the matrix unit. For $t\ne 0$ this is a transvection: it fixes the hyperplane with $j$th coordinate zero pointwise, has determinant one and inverse $E_{ij}(-t)$. We show these elementary transvection matrices generate $SL(n,F)$ over any field.

Let $M\in SL(n,F)$. Row addition is left multiplication by an $E_{ij}(t)$. If the current diagonal pivot is zero, a nonzero entry occurs below it in the same remaining column, since the remaining square block is invertible; adding that row makes the pivot nonzero. Row additions then clear the entries below the pivot. Repeating produces an upper-triangular matrix with nonzero diagonal. Clearing the entries above the diagonal, starting from the last column, gives a diagonal matrix $D=\mathrm{diag}(a_1,\dots ,a_n)$ with ${\prod }_i{a}_i=1$. Thus it remains to realize determinant-one diagonal matrices using elementary transvections.

In a two-coordinate block putDirect multiplication givesThe diagonal matrix $D$ is the product, for $i=1,\dots ,n-1$, of these blocks on coordinates $i,n$, with entries $a_i,a_i^{-1}$. Their final $n$th entry is ${\prod }_{i<n}{a}_i^{-1}=a_n$. Therefore $D$, and hence $M$, is a product of elementary transvections. All row operations used are invertible transvections, and no division except by a nonzero pivot or $a_i$ has been made. This provesThe identity element $E_{ij}(0)$ can be omitted from the generating set. This is transvections generate the special linear group; ordinary row scaling alone would not prove it, because such a scaling can leave $SL(n,F)$.

Let $V=F^n$ and let $G=PSL(n,F)$ act on its one-dimensional subspaces. First identify the kernel of a group homomorphism in $SL(n,F)$. A matrix fixing every coordinate line is diagonal, and fixing every line $⟨e_i+e_j⟩$ forces all diagonal entries equal. Thus the projective kernel of a group homomorphism is $\{\lambda I:{\lambda }^n=1\}$. These scalars are central. Conversely a central matrix commutes with all $E_{ij}(1)$: comparing entries in $M{e}_{ij}=e_{ij}M$ shows all off-diagonal entries of $M$ vanish and all diagonal entries agree. Therefore the kernel of a group homomorphism is exactly $Z(SL(n,F))$. This proves scalar kernel of the projective linear action and faithfulness of the induced $G$-action.

The action is doubly transitive. Given two ordered pairs of distinct lines, choose two independent representatives of each pair and extend them to bases. A linear map taking the first basis to the second sends the desired lines correctly. If its determinant is $d\ne 1$, postcompose with a diagonal map in the target basis that scales its first basis vector by $d^{-1}$ and fixes all the others. It preserves the two target lines and corrects the determinant. Thus an element of $SL(n,F)$ sends either ordered pair to the other. Double transitivity implies primitivity: a block containing two distinct points must contain every point by the transitive point-stabilizer action.

For $\alpha =⟨e_1⟩$, considerProducts add the functionals, so $U$ is abelian. If $g$ fixes $\alpha$, write $g{e}_1=a{e}_1$. Then $g(I+e_1\phi )g^{-1}=I+e_1(a\phi g^{-1})$, and the new functional still vanishes on $e_1$. Thus the image $A$ of $U$ in $G$ is normal in the point stabilizer. Its conjugates contain all elementary transvections: changing the center line $⟨e_1⟩$ to $⟨e_i⟩$ gives every map $I+e_i\psi$ with $\psi (e_i)=0$, including $E_{ij}(t)$. Part (b) then shows that those conjugates generate $G$.

It remains to prove perfectness, not assume it. Use $[x,y]=x^{-1}{y}^{-1}xy$. For $n\ge 3$, distinct $i,j,k$ giveEvery generator is therefore a group commutator, so $SL(n,F{)}^{\prime }=SL(n,F)$. For $n=2$ and $|F|>3$, choose $a\in F^{\times }$ with $a^2\ne 1$, possible because a quadratic polynomial has at most two roots. With $D=\mathrm{diag}(a,a^{-1})$,Both coefficients are nonzero, so every upper and lower elementary transvection is again a group commutator. Part (b) proves perfectness in this case as well. Quotients of perfect groups are perfect, hence $G^{\prime }=G$ in exactly the stated range.

Apply part (a) to the faithful primitive $G$-action and its abelian normal point-stabilizer subgroup $A$. Every nontrivial normal subgroup contains $G^{\prime }=G$, so it is the whole group. A transvection is not scalar, showing $G\ne 1$. We have provedThe proof works for infinite fields too and uses no finite-order count. The excluded two-dimensional fields give the familiar soluble exceptions $PSL(2,2)\cong S_3$ and $PSL(2,3)\cong A_4$.

Write $x_{ij}(t)=I+t{E}_{ij}$ for $i\ne j$ and $t\in {\mathbb{F}}_q$. When $t\ne 0$ these are transvections, and their inverses are $x_{ij}(-t)$. Row addition by such an elementary transvection matrix preserves the determinant. It also realizes signed interchanges and compensating diagonal scalings: in any two coordinate positions,Thus elimination can reduce any invertible matrix to diagonal form using only row additions and signed interchanges, all products of transvections. For a diagonal matrix with determinant one, $\mathrm{diag}(d_1,\dots ,d_n)$, use the displayed two-coordinate diagonal matrices in positions $(i,n)$ with parameter $d_i$, for $1\le i<n$. Their product has diagonal entries $d_1,\dots ,d_{n-1},{\prod }_{i<n}{d}_i^{-1}=d_n$. Hence the residual diagonal matrix is also a product of transvections, and reversing elimination provesThis proves transvections generate the special linear group, including the small fields.

For perfectness, use the convention $[x,y]=xy{x}^{-1}{y}^{-1}$ in the following calculations. If $n\ge 3$ and $i,j,k$ are distinct, direct matrix multiplication givesTaking $b=1$ puts every elementary generator in the commutator subgroup. Thus $S{L}_n(q)$ is perfect. When $n=2$ and $q>3$, choose $s\in {\mathbb{F}}_q^{\times }$ with $s^2\ne 1$; such an element exists because a degree-two polynomial has at most two roots and $q-1>2$. For $D=\mathrm{diag}(s,s^{-1})$,Both coefficients are nonzero, so varying $t$ supplies every upper and lower elementary generator as a commutator. ThereforeThe exceptions really fail: $S{L}_2(2)\cong S_3$ has a nontrivial sign quotient, and $S{L}_2(3)$ has quotient $PS{L}_2(3)\cong A_4$, which has the nontrivial Abelian quotient $A_4/V_4\cong C_3$. For the latter identification, the faithful projective action embeds the order-twelve group $PS{L}_2(3)$ as an index-two subgroup of $S_4$, necessarily $A_4$.

The Iwasawa simplicity lemma states that, in a faithful primitive group action, if a point stabilizer has an Abelian normal subgroup whose conjugates generate the whole group, then every nontrivial normal subgroup contains the commutator subgroup. In particular a nontrivial perfect group with this property is simple. The lemma is applied here, as allowed, without assuming the desired simplicity theorem.

Let $S{L}_n(q)$ act on the one-dimensional subspaces of ${\mathbb{F}}_q^n$. This action is two-transitive: a linear map taking two specified distinct lines to two others can be made determinant one by rescaling one of the target basis vectors, without changing either target line. Its kernel consists exactly of scalar matrices. Indeed fixing all lines makes a matrix diagonal in any fixed basis, and fixing the lines through $e_i+e_j$ makes all its diagonal entries equal. The scalar matrices in $S{L}_n(q)$ are $aI$ with $a^n=1$. They also form its group center, since a central matrix commutes with every elementary generator, forcing all off-diagonal entries to vanish and all diagonal entries to agree. Their number is $\gcd (n,q-1)$. Consequently the quotient $PS{L}_n(q)$ acts faithfully and primitively.

Fix a line $L=⟨v⟩$ and considerHere $vf$ is the rank-at-most-one map $z\mapsto f(z)v$. These matrices have determinant one, and $(vf)(vh)=0$, so $T_L$ is Abelian under addition of the functionals. If $gL=L$, then $gv=av$ and $g(I+vf)g^{-1}=I+av(f{g}^{-1})\in T_L$. Thus $T_L$ is normal in the line stabilizer. Its image in $PS{L}_n(q)$ is a nontrivial Abelian normal subgroup of the point stabilizer. Conjugates of these groups include every elementary transvection, hence generate $PS{L}_n(q)$. Perfectness passes to a quotient. The Iwasawa simplicity lemma therefore givesIn those exceptions the projective groups are $S_3$ and $A_4$, as above.

For $q=4$, counting columns gives $|S{L}_2(4)|=4(4^2-1)=60$. Its scalar group center is trivial, so $PS{L}_2(4)$ has order sixty. Its faithful action on the five points of the projective line embeds it in $S_5$. Because it is perfect, its sign homomorphism is trivial and its image lies in $A_5$. Equal orders then giveFor $q=5$, use an explicit subgroup to obtain a five-point action. In $S{L}_2(5)$ setAll three have determinant one. Matrix multiplication gives $i^2=j^2=-I$, $ij=-ji$, and $r^3=I$. The eight distinct matrices $\{\pm I,\pm i,\pm j,\pm ij\}$ form a quaternion group. Moreover,Thus $r$ normalizes that quaternion group and has order three. Their generated subgroup has order $24$ and contains the scalar group center $\{\pm I\}$. Its image in $PS{L}_2(5)$ has order twelve, hence index five since $|PS{L}_2(5)|=5(25-1)/2=60$. The resulting coset action is nontrivial and, by the simplicity already proved, faithful. Perfectness again puts its image in $A_5$, and equal orders implyThis is the quaternion construction of an index-five subgroup of PSL2 over F5; it avoids assuming a classification of groups of order sixty.

For the last comparison, the order of a general linear group over a finite field comes from choosing independent columns:The determinant map onto ${\mathbb{F}}_q^{\times }$ has kernel $S{L}_n(q)$, and the scalar quotient divides by $\gcd (n,q-1)$. HenceThe highest power of two dividing either order is $64$. Their Sylow $2$-subgroups are respectively the upper unitriangular groups $U_4(2)$ of order $2^6$ and the image of $U_3(4)$ of order $4^3$. The latter image is isomorphic to $U_3(4)$ because its intersection with the odd-order scalar group center is trivial.

The center of an upper unitriangular group can be computed directly here. Commuting $X\in U_4(2)$ with $I+E_{12}$ forces $X_{23}=X_{24}=0$; commuting with $I+E_{23}$ forces $X_{12}=X_{34}=0$; commuting with $I+E_{34}$ forces $X_{13}=0$. Only $X_{14}$ can remain, and those matrices commute with the whole group. Thus $Z(U_4(2))=\{I+t{E}_{14}:t\in {\mathbb{F}}_2\}$ has order two. In $U_3(4)$, commutation with $I+E_{12}$ and $I+E_{23}$ similarly forces $X_{23}=X_{12}=0$, leaving the arbitrary entry $X_{13}\in {\mathbb{F}}_4$. Its group center has order four. An isomorphism of the whole groups would send a Sylow subgroup to a Sylow subgroup and preserve its group center order, which is impossible. Therefore

Use finitely generated right projective modules throughout. The group K0 of a ring is the Grothendieck group of their direct-sum monoid, with $[P\oplus Q]=[P]+[Q]$. The group K1 of a ring is ${\mathrm{GL}}_{\infty }(A)/E_{\infty }(A)$, equivalently the abelianization of the stable general linear group. Stabilization is $g\mapsto \mathrm{diag}(g,1)$; $E_{\infty }$ is generated by elementary transvections. The stable elementary subgroup is the commutator subgroup: elementary transvections are commutators using a third index, and stable block interchange and elementary row identities make the quotient abelian.

Write $A_0$ for the lower-right ring, printed as $A^{\prime }$ in the source. A Milnor square has unital commuting maps, is a pullbackand has at least one of $j_1,j_2$ surjective. Interchange the two sides if necessary so that $j_1$ is surjective.

We first establish the patching fact used in the proof. For projective $P_i$ and an isomorphism $\varphi :P_1\otimes A_0\to P_2\otimes A_0$, setFor free modules denote the result by $M_g$. The block matrix identityexpresses $\mathrm{diag}(g,g^{-1})$ as elementary matrices. Their entries lift through $j_1$, and their elementary lifts remain invertible. A change of basis on the first side then trivializes this gluing. Thus $M_g\oplus M_{g^{-1}}\cong A^{2r}$, proving that $M_g$ is finitely generated projective. Notice that we lifted elementary matrices, not arbitrary invertible matrices.

For general $P_1,P_2$, choose $P_1\oplus Q_1\cong A_1^r$ and $P_2\oplus Q_2\cong A_2^s$. The reductions of $Q_1\oplus A_1^s$ and $Q_2\oplus A_2^r$ are isomorphic: insert the common reduced $P_i$ and reorder the two complements. Patch them by such an isomorphism. The sum with $M(P_1,P_2,\varphi )$ now patches two free modules of equal size, so is projective by the free case, making each summand projective. The same free change of basis followed by these direct-summand projections shows that extension to $A_i$ recovers $P_i$. Conversely, write an $A$-projective as the image of an idempotent over a finite free module. Since matrices over the pullback are compatible pairs of matrices, applying the compatible idempotents to the free pullback reconstructs this projective from its two reductions. Compatible module isomorphisms patch as well. This proves Milnor patching of projective modules in the form needed below.

The diagonal maps in the sequence are extension of scalars; the two maps to $A_0$ are differences $j_{1\ast }-j_{2\ast }$. Define the connecting map byIt is unchanged by stabilization or elementary multiplication: an elementary factor lifts through $j_1$ and changes a free basis in the gluing. The identity $\mathrm{diag}(g,h)=\mathrm{diag}(gh,I)\mathrm{diag}(h^{-1},h)$ and the preceding elementary factorization show $M_g\oplus M_h\cong M_{gh}\oplus A^r$ after matching sizes. Thus $\partial$ is an additive homomorphism on $K_1(A_0)$.

Exactness at $K_1(A_1)\oplus K_1(A_2)$. If the two reduced classes agree, choose stabilized representatives $g_1,g_2$ of the same size. Their reductions differ by an elementary matrix $e$ over $A_0$. Lift $e$ to an elementary matrix over $A_1$ and adjust $g_1$ without changing its $K_1$ class. Its reduction now equals that of $g_2$. The compatible pair and its compatible inverse give an invertible matrix over $A$, so the original pair of classes is in the diagonal image. The reverse containment follows from commutativity of the square.

Exactness at $K_1(A_0)$. A gluing obtained from a change of basis over either upper ring patches a free module, so the difference-map image is killed by $\partial$. Conversely, $\partial [g]=0$ means that $M_g$ becomes free after adding a common projective summand. Add a complement of that summand to obtain an actual isomorphism $M_g\oplus A^s\cong A^{r+s}$. Its two local bases express $g\oplus I_s=j_2(v_2)j_1(v_1{)}^{-1}$. Therefore $[g]=j_{2\ast }[v_2]-j_{1\ast }[v_1]$, which lies in the image of the difference map. This uses stable equality of projective classes, not cancellation of arbitrary projectives.

Exactness at $K_0(A)$. Connecting classes reduce to zero on both sides. Conversely, write a class in the kernel as $[P]-[A^r]$ by adding a complement to its negative projective term. Its reductions have the same classes as $A_i^r$. Equality in $K_0(A_i)$ becomes an isomorphism after a finite common free summand, by taking a complement to the auxiliary projective. Use one sufficiently large free summand on both sides. The stabilized $P$ is then locally free with gluing $g$, so the original class is $[M_g]-[A^{r+s}]=\partial [g]$.

Exactness at $K_0(A_1)\oplus K_0(A_2)$. Write a kernel pair as $([P_1]-[A_1^r],[P_2]-[A_2^r])$, using complements and padding to make the free sizes equal. Their reduced positive classes agree. Adding a common finite free summand makes those reductions isomorphic. Patch the resulting projectives, and subtract the matching free $A$-module. Its restrictions are the desired pair. The reverse containment is immediate because both reductions to $A_0$ agree.

These arguments prove the Mayer-Vietoris sequence in algebraic K-theoryNo zero is inserted at either end; this fragment does not assert injectivity at the first term or surjectivity at the last.

For the group ring of the cyclic group, take the cyclotomic Milnor squareThe maps send the cyclic generator to $1$ and to $\zeta$; the lower map sets $\zeta =1$, and the right map reduces modulo $p$. Since ${\Phi }_p(1)=p$, $\mathbb{Z}[\zeta ]/(\zeta -1)={\mathbb{F}}_p$. To verify the pullback, represent an element by a polynomial of degree at most $p-1$. If both evaluations vanish, it is a constant multiple of the minimal polynomial of an algebraic element ${\Phi }_p$, and evaluation at one forces that constant to be zero. Conversely, represent $b\in \mathbb{Z}[\zeta ]$ by $f$ of degree below $p-1$. For a compatible integer $a$, $a-f(1)$ is divisible by $p$; replacing $f$ by $f+[(a-f(1))/p]{\Phi }_p$ yields the pair $(a,b)$. The right map is surjective, so this is a Milnor square.

The hinted cyclotomic units are $u_k=1+\zeta +\cdots +{\zeta }^{k-1}$. If $kl\equiv 1(\mathrm{mod}p)$ with $1\le l<p$, the geometric sum ${\sum }_{j=0}^{l-1}{\zeta }^{kj}$ is their inverse, because ${\zeta }^{kl}=\zeta$. Their residues are $k\in {\mathbb{F}}_p^{\times }$, covering every unit of the common field. Gaussian elimination identifies $K_1({\mathbb{F}}_p)$ with ${\mathbb{F}}_p^{\times }$ by determinant. Thus $K_1(\mathbb{Z}[\zeta ])\to K_1({\mathbb{F}}_p)$ is surjective, and the connecting map vanishes.

The remaining exact sequence injects $K_0(\mathbb{Z}[C_p])$ into $\mathbb{Z}\oplus K_0(\mathbb{Z}[\zeta ])$, with image $\{(r,x):r=\mathrm{rank}x\}$. Here $K_0(\mathbb{Z})=K_0({\mathbb{F}}_p)=\mathbb{Z}$, and reduction of a projective over the integral domain $\mathbb{Z}[\zeta ]$ has dimension its rank: localize at $(\zeta -1)$, where the projective is free, and compare its residue-field dimension with its fraction-field dimension. Projection onto the second factor is therefore a bijection on this kernel. Hence extension along the cyclotomic evaluation givesThe essential step is reduction of cyclotomic units at the ramified prime; surjectivity of a ring map alone would not imply surjectivity on $K_1$.

For $i\ne j$, put $x_{ij}(a)=I+a{E}_{ij}$. When $a\ne 0$ this is a transvection: it fixes the hyperplane $v_j=0$ pointwise and has rank-one difference from the identity. Its inverse is $x_{ij}(-a)$ and its determinant is one.

Left multiplication performs a row addition. For an invertible matrix, bring a nonzero entry into a pivot position by a signed interchange, then clear its column by row additions. A signed interchange on two coordinates is itself a product of row additions, becauseRepeating the pivot construction and then clearing above the pivots reduces any determinant-one matrix to a diagonal matrix of determinant one. FinallyA determinant-one diagonal matrix is a product of such two-coordinate matrices, using the last coordinate to compensate each of the first $n-1$ diagonal entries. Each displayed matrix is a product of elementary transvection matrices. Hence reversing the elimination provesThis argument works over every field, including characteristic two.

Let $S=S{L}_n(F)$ and let $Z$ be its scalar center. Its action on the lines of $F^n$ has kernel exactly $Z$: a linear map fixing every line is diagonal in a basis, and fixing the lines spanned by $e_i+e_j$ makes all its diagonal entries equal. The action is two-transitive. Indeed an ordered pair of distinct lines can be represented by two independent vectors and extended to a basis. A linear map between two such bases can be adjusted to determinant one by scaling one target basis vector; this preserves both target lines. Therefore $S/Z$ has a faithful primitive group action.

Fix $L=Fv$. The mapsform an abelian subgroup: all products of the rank-one terms vanish. It is normal in the line stabilizer, since conjugation replaces $v$ by a scalar multiple and transports $f$. Its conjugates contain every transvection, and hence generate $S$ by 4(b). The same statements hold for its image in $S/Z$.

It remains to prove perfectness rather than assume it. Use the group commutator convention $[x,y]=xy{x}^{-1}{y}^{-1}$. If $n\ge 3$, distinct $i,j,k$ giveEvery elementary generator is consequently in $S^{\prime }$. If $n=2$ and $|F|>3$, choose $a\in F^{\ast }$ with $a^2\ne 1$. There are at most two roots of $a^2=1$, so such a choice exists. For $h=\mathrm{diag}(a,a^{-1})$,Varying $b$ gives every upper transvection; conjugating by $w(1)$ gives every lower one. Again $S^{\prime }=S$, and its projective quotient is a perfect group.

Apply 4(a) to this faithful primitive projective action. Every nontrivial normal subgroup contains the commutator subgroup, which is the whole quotient. ThusThe restriction is real: $PS{L}_2(2)\cong S_3$ and $PS{L}_2(3)\cong A_4$ are not simple.

Replace $F^n$ by a symplectic vector space $(V,B)$ of dimension $2m$, and elementary row operations by symplectic transvectionsThe cross terms in $B(Tx,Ty)$ cancel, and $B(v,v)=0$, so these maps preserve $B$. Their inverses are $T_{v,-c}$.

They generate $Sp(V)$ by the following symplectic basis elimination. If $B(u,v)\ne 0$, a single transvection with direction $v-u$ and parameter $1/B(u,v)$ sends $u$ to $v$. If $B(u,v)=0$ and $u\ne v$, choose $w$ with both $B(u,w)$ and $B(w,v)$ nonzero: the union of two proper hyperplanes does not exhaust a vector space, even over ${\mathbb{F}}_2$. Two transvections then suffice. Thus one can make a form-preserving linear map fix the first basis vector $e$. Next align its partner $f$, while keeping $e$ fixed. Two partners have difference in $e^{\perp }$, so the preceding single-transvection construction uses a direction in $e^{\perp }$ whenever their mutual pairing is nonzero. If it is zero, the intermediate partner $f+ae$, $a\ne 0$, has nonzero pairing with both. All these transvections fix $e$. Once both $e,f$ are fixed, recurse on their nondegenerate symplectic orthogonal complement. This proves generation over every field.

Use the action on the points of projective space, whose kernel is the scalar center $Z=\{\lambda I:{\lambda }^2=1\}$. For $m=1$ this is the $S{L}_2$ case. For $m\ge 2$ the action is primitive, although it is generally not two-transitive. The stabilizer of a line $L$ has two orbits on other lines: those orthogonal to $L$ and those not orthogonal to $L$. This follows by extending a hyperbolic pair, or an independent isotropic pair, to a symplectic basis. A block containing $L$ and a second line $M$ contains the entire corresponding orbit of the stabilizer of $L$. The stabilizer of $M$ then supplies a line of the other type: choose a vector orthogonal to one of $L,M$ and not to the other. Nondegeneracy and independence guarantee such a vector, also in dimension four. Hence the block contains both orbits and is the whole set.

The transvections along $L$ form an abelian normal subgroup of its stabilizer; their conjugates generate the group. Thus the Iwasawa simplicity lemma applies as soon as perfectness is established. For $|F|>3$, every transvection lies in an embedded $S{L}_2(F)$ acting on a hyperbolic plane, and 4(c) proves that embedded group perfect.

For the remaining fields, the perfectness of finite symplectic groups has a short direct proof. Over ${\mathbb{F}}_3$ with $m\ge 2$, all parameter-one transvections have the same class $t$ in the abelianization, and $3t=0$. In an isotropic plane with basis $u,v$, the four directions $u,v,u+v,u-v$ give commuting transvections whose product is $1$: their rank-one terms add to zero, sinceover ${\mathbb{F}}_3$. Thus $4t=0$, giving $t=0$. Parameter-two transvections are inverses, so all generators vanish in the abelianization. Over ${\mathbb{F}}_2$ with $m\ge 3$, use the seven nonzero vectors of an isotropic three-space. Their commuting transvections multiply to $1$, since each diagonal coordinate of ${\sum }_{0\ne v}v{v}^{\mathsf{T}}$ occurs four times and each off-diagonal coordinate twice. Their common abelianized class satisfies $2t=7t=0$, and again the group is perfect.

Consequently the projective symplectic group is simple, with the necessary exceptionsThe first two are the linear exceptions; the last is $S{p}_4(2)\cong S_6$, proved in 6(c). The word “projective” matters: the unquotiented symplectic group may have the nontrivial scalar center $\{I,-I\}$.