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Choose a normal subgroup $N$ of prime order $p\in \{5,7\}$ as in ii, and choose a subgroup $H$ of the other prime order $q$. Their intersection is trivial by Lagrange's theorem. Both are cyclic groups, and ii gives $|\mathrm{Aut}(N)|=p-1$.

The image of a group homomorphism $\pi :H\to \mathrm{Aut}(N)$ from iii has order dividing $q$, because its kernel of a group homomorphism partitions $H$ into equally sized cosets. Its order also divides $p-1$ by Lagrange's theorem. Butso $\pi$ is trivial in either case. By iii, $⟨N,H⟩\cong C_p\times C_q$ has order $pq=35$ and therefore equals $G$.

If $n$ and $h$ generate $N$ and $H$, the element $(n,h)$ has order $\mathrm{lcm}(p,q)=pq=35$, since its $k$th power is the identity exactly when both $p$ and $q$ divide $k$. Thus this direct product of groups is cyclic, and

Definitions. A group homomorphism $\phi :G_1\to G_2$ preserves multiplication, $\phi (xy)=\phi (x)\phi (y)$. A group isomorphism is a bijective group homomorphism. The kernel of a group homomorphism isPreserving multiplication implies $\phi (e_1)=e_2$ and $\phi (x^{-1})=\phi (x{)}^{-1}$: the first follows by cancelling $\phi (e_1)$ from its square, and the second follows by applying $\phi$ to $x{x}^{-1}=e_1$.

Kernel subgroup. The identity belongs to $K$. For $x,y\in K$, $\phi (xy)=e_2{e}_2=e_2$, so $xy\in K$. Also $\phi (x^{-1})=e_2^{-1}=e_2$, so $x^{-1}\in K$. Thus $\boxed{K\le G_1}$.

Normality. For $g\in G_1$ and $k\in K$,Hence every $gk{g}^{-1}$ is again in $K$, proving $\boxed{K⊴G_1}$. Applying the same statement with $g^{-1}$ gives equality $gK{g}^{-1}=K$, and therefore $gK=Kg$.

Group of cosets. Define the product on the left cosets byTo check that this does not depend on representatives, replace $a,b$ by $a{k}_1,b{k}_2$. Thenbecause normality puts $b^{-1}{k}_{1}b$ in $K$. Thus the product is well defined. It is associative because multiplication in $G_1$ is associative; its identity is $K=e_{1}K$, and the inverse of $aK$ is $a^{-1}K$. Consequently the cosets form the quotient group $\boxed{G_1/K}$.

The map $aK\mapsto \phi (a)$ is well defined and gives an isomorphism $G_1/K\cong \mathrm{im}\phi$. If the original mapping is onto $G_2$, this becomes $G_1/K\cong G_2$. No surjectivity is needed for the kernel of a group homomorphism to be a normal subgroup or for the quotient group to exist.

A normal subgroup $H$ satisfies $gH{g}^{-1}=H$ for every $g\in G$. A group homomorphism preserves multiplication, $\Phi (ab)=\Phi (a)\Phi (b)$. It follows that $\Phi (e_1)=e_2$ and $\Phi (g^{-1})=\Phi (g{)}^{-1}$.

The kernel of a group homomorphism $K=\{g:\Phi (g)=e_2\}$ contains the identity and satisfies $a{b}^{-1}\in K$ for $a,b\in K$, so is a subgroup. For $k\in K$, $\Phi (gk{g}^{-1})=\Phi (g)e_2\Phi (g{)}^{-1}=e_2$. Thus $gK{g}^{-1}\subseteq K$, and applying the same inclusion to $g^{-1}$ gives equality. Hence the kernel is a normal subgroup.

The product $f={\prod }_{i<j}(x_i-x_j)$ is a Vandermonde determinant up to a fixed overall sign. A transposition exchanging $x_p,x_q$ reverses their mutual factor. Every other factor involving one of these variables is paired with the corresponding factor involving the other; those paired products are unchanged. ThereforeIf a permutation has decompositions into $r$ and $s$ transpositions, they act as $(-1{)}^{r}f$ and $(-1{)}^{s}f$. This polynomial is nonzero, for example at pairwise distinct real inputs, so the signs must agree. Thus all decompositions have the same parity, and an even decomposition forces every decomposition to be even. This defines the sign of a permutation; its kernel is the alternating group, linking the parity argument to the normal-kernel result above.

A group homomorphism satisfies $\phi (g_1{g}_2)=\phi (g_1)\phi (g_2)$; a bijective one is a group isomorphism. Its kernel of a group homomorphism is $K=\{g:\phi (g)=1_H\}$. It is a subgroup, and for $k\in K$, $g\in G$,Thus $gK{g}^{-1}\subseteq K$, and applying the same argument to $g^{-1}$ gives equality. Therefore $K$ is a normal subgroup.

Write $B_n=\left(\begin{array}{cc}1& n\\ 0& 1\end{array}\right)$. Direct matrix multiplication gives $B_n{B}_m=B_{n+m}$, $B_0=I$ and $B_n^{-1}=B_{-n}$. Consequently these matrices form the integer shear group, a subgroup of $GL(2,\mathbb{R})$. It is an infinite cyclic group, generated by $B_1$, since $B_1^n=B_n$ and no nonzero integer power is the identity.

The map $n\mapsto B_n$ is a bijective group homomorphism from the additive group of integers, proving the required group isomorphism. There are proper subgroups: for example $\{B_{2n}:n\in \mathbb{Z}\}$ omits $B_1$. In fact every subgroup is $\{B_{mn}:n\in \mathbb{Z}\}$ for an integer $m\ge 0$, by the classification of subgroups of an infinite cyclic group; $m=0$ gives the trivial subgroup and $m=1$ the whole group.

The kernel of a group homomorphism is $K=\{g:\theta (g)=1_H\}$. It is a subgroup, since the identity maps to the identity and $\theta (k_1{k}_2^{-1})=1_H$ for $k_1,k_2\in K$. Alsoso $gK{g}^{-1}\subseteq K$. Applying the same inclusion with $g^{-1}$ gives equality, proving that $K$ is a normal subgroup.

The map $gK\mapsto \theta (g)$ is well defined, multiplicative and injective, with image $\mathrm{im}\theta$. Thus the first isomorphism theorem givesThe quotient is isomorphic to all of $H$ precisely when the original homomorphism is surjective.

For the general linear group, the determinant gives a homomorphism $\det :\mathrm{GL}(n,\mathbb{R})\to {\mathbb{R}}^{\times }$, since $\det (AB)=\det A\det B$. Its kernel is the special linear group $\mathrm{SL}(n,\mathbb{R})$, so that subgroup is normal. Every nonzero real $t$ is the determinant of $\mathrm{diag}(t,1,\dots ,1)$ for $n\ge 1$, making the map surjective. Thereforewhere the operation on the right is multiplication.

Let $B=\left(\begin{array}{cc}0& b\\ c& 0\end{array}\right)$ be the lower block. ThenThe defining relation $a^4=I$ requires $T(a{)}^4=I_3$, so for real constants $bc=\pm 1$. Both possibilities ensure invertibility.

If $bc=-1$, the generator matrix has order four, because its square is not the identity while its fourth power is. Its four powers are distinct, giving a faithful representation. If $bc=1$, its square is the identity but it is not itself the identity; its lower diagonal entries are zero. The image then has order two and the kernel of a group homomorphism is $\{I,a^2\}$. Thus the faithfulness criterion for a block representation of the cyclic group of order four givesOther parameter values give no representation of $Z_4$, not additional unfaithful cases; $bc=0$ also makes the assigned matrix singular. No orthogonality of these matrices was required.

For the final construction takeMultiplication adds exponents, so $D(a^r)D(a^s)=D(a^{r+s})$ and $D(a{)}^4=I_2$. The representing matrix is the identity exactly for even exponents. Therefore this is a two-dimensional representation with kernel $\{I,a^2\}$, as required. It factors through the quotient cyclic group of order two.

The image of a group homomorphism contains $e_{G_2}=\varphi (e_{G_1})$. If $u=\varphi (g)$ and $v=\varphi (h)$ lie in the image, then $u{v}^{-1}=\varphi (g{h}^{-1})$ also lies there. The subgroup test proves $\mathrm{im}\varphi \le G_2$.

The kernel of a group homomorphism is $K=\{g:\varphi (g)=e_{G_2}\}$. It is nonempty and closed under $k{h}^{-1}$ for $k,h\in K$, so it is a subgroup. For any $g\in G_1$ and $k\in K$,Thus $gK{g}^{-1}\subseteq K$. Replacing $g$ by $g^{-1}$ gives the reverse inclusion, provingFinally, if $K=\{e\}$ and $\varphi (g)=\varphi (h)$, then $\varphi (h^{-1}g)=e$. Thus $h^{-1}g\in K$ forces $g=h$, establishing injectivity. Conversely an injective homomorphism has trivial kernel, soNone of these arguments requires either group to be finite.

The kernel of a group homomorphism isThe homomorphism sends identity to identity and inverses to inverses. Thus $e_G\in K$, and if $k_1,k_2\in K$, then $\Phi (k_1{k}_2^{-1})=e_{G^{\prime }}$, so $K$ is a subgroup by the subgroup test.

For any $g\in G$ and $k\in K$,Hence $gK{g}^{-1}\subseteq K$. Applying the same inclusion with $g^{-1}$ gives equality, provingThis normality argument does not require either group to be finite.

A group homomorphism $\varphi :H\to G$ is a map preserving multiplication:It necessarily sends $e_H$ to $e_G$ and sends inverses to inverses, by applying this relation to $e_H^2$ and $h{h}^{-1}$. Its kernel of a group homomorphism isThe kernel is a normal subgroup of the domain: if $k\in \ker \varphi$ and $h\in H$, then $\varphi (hk{h}^{-1})=\varphi (h)e_G\varphi (h{)}^{-1}=e_G$. The kernel’s identity condition refers to the identity in the target group.

A member of $H$ is $A(t,b,t)$. The multiplication law from part (a) becomesIt stays in $H$ and is unchanged when the two factors are exchanged, since $ts=st$. The identity lies in $H$, and $A(t,b,t{)}^{-1}=A(-t,t^2-b,-t)$ also lies in $H$. Thus $H$ is an Abelian subgroup, namely the Abelian plane subgroup of a finite Heisenberg group in this coordinate convention.

There are five choices each for $t,b$, soThis verifies Lagrange's theorem: the subgroup order divides the group order, and the number of distinct left cosets is five.

Explicitly, the map $\phi (A(a,b,c))=a-c$ is a surjective group homomorphism to the additive group of ${\mathbb{F}}_5$, by the product formula. Its kernel of a group homomorphism is $H$. Thus $H$ is even a normal subgroup; the five left cosets are the level sets $a-c=d$ for $d=0,1,2,3,4$, with representatives $A(d,0,0)$. Each contains 25 elements, providing a direct count as well.

The kernel of a group homomorphism found in part (b) is the translation subgroupIt is a subgroup because $T_c{T}_d=T_{c+d}$, $T_0=I$ and $T_c^{-1}=T_{-c}$. It is nontrivial since $T_1\ne I$, and proper since, for example, $M(2,0)\notin N$.

Every kernel is a normal subgroup: if $n\in \ker f$, then $f(gn{g}^{-1})=f(g)ef(g{)}^{-1}=e$. Applying the same argument with $g^{-1}$ gives equality of the conjugate subgroup and the kernel. Here one can also check invariance under conjugation directly:Multiplication by nonzero $a$ permutes all $c\in \mathbb{C}$, so $M(a,b)NM(a,b{)}^{-1}=N$. Thus $N$ is a nontrivial proper normal subgroup, isomorphic to the additive group $(\mathbb{C},+)$.

The sign homomorphism $\mathrm{sgn}:S_n\to \{+1,-1\}$ records parity of a permutation. The kernel of a group homomorphism in this case is the alternating group $A_n$, consisting of even permutations. For $n\ge 2$ it has index two in the symmetric group and order $n!/2$.

The cycle type is the multiset of lengths of the disjoint cycles, including fixed points if desired. For $A_4$ the complete count isA single transposition and a four-cycle are odd and are excluded. Thus

A group homomorphism $f:G\to H$ preserves the operation: $f(ab)=f(a)f(b)$ for all $a,b\in G$. A group isomorphism is a bijective group homomorphism; its inverse also preserves the operation. The kernel of a group homomorphism is $\ker f=\{g\in G:f(g)=e_H\}$, a normal subgroup of $G$.

The three nonzero vectors in ${\mathbb{F}}_2^2$ are $e_1$, $e_2$ and $e_1+e_2$. An invertible linear map permutes this three-element set, so part (b) gives a group homomorphismIf a matrix acts trivially on all three vectors, it fixes both basis vectors $e_1,e_2$, so it is the identity matrix. Thus the kernel of a group homomorphism is trivial and $\rho$ is injective.

To count the domain, its first column may be any of the three nonzero vectors. The second column must be outside the two-element span of the first, leaving two choices. HenceAn injection between these finite sets of equal size is surjective, proving

Define the group homomorphismBy part (b)(i), its kernel of a group homomorphism isIt is surjective. Indeed, choose $r,s\in \mathbb{Z}$ with $rd+sh=1$ by Bézout's identity. For any prescribed classes $u+d\mathbb{Z}$ and $v+h\mathbb{Z}$, the integersatisfies $t\equiv u(\mathrm{mod}d)$ and $t\equiv v(\mathrm{mod}h)$. The first isomorphism theorem for groups therefore givesThis proves the required isomorphism and is the Chinese remainder theorem for coprime integers. It also shows that $(1+d\mathbb{Z},1+h\mathbb{Z})$ generates the direct sum of modules when these additive groups are viewed as $\mathbb{Z}$-modules.

For an odd prime $p$, a nonzero congruence class $a$ is a quadratic residue when $a\equiv x^2(\mathrm{mod}p)$ for some integer $x$. Here "quadratic residue" excludes zero: zero is a square of a number, but is neither a quadratic residue nor a quadratic nonresidue under this convention. Some treatments include zero among quadratic residues; the distinction matters when counting them or discussing their subgroup.^[ref]

Modulo seven, squaring the six nonzero congruence classes givesThus the set of quadratic residues is $Q_7=\{1,2,4\}$, and the quadratic nonresidues are $\{3,5,6\}$. In particular, $x^2\equiv 2(\mathrm{mod}7)$ has exactly the two solutions $x\equiv 3,4(\mathrm{mod}7)$: $3^2=9\equiv 2$ and $4^2=16\equiv 2$. Modulo eleven the quadratic residues are instead $Q_{11}=\{1,3,4,5,9\}$. Two is therefore a quadratic residue modulo seven and a quadratic nonresidue modulo eleven: being a quadratic residue depends on the modulus.

Why do the square roots come in pairs? If $x^2\equiv y^2(\mathrm{mod}p)$, the prime number $p$ divides $(x-y)(x+y)$, so it divides one factor: $y\equiv x$ or $y\equiv -x(\mathrm{mod}p)$. For nonzero $x$ these are distinct, because $p$ is an odd prime. Every nonempty nonzero fiber of a function of $s(x)=x^2$ therefore contains exactly two inputs. Pairing the $p-1$ nonzero congruence classes gives exactly $(p-1)/2$ quadratic residues and $(p-1)/2$ quadratic nonresidues. Including zero gives $(p+1)/2$ different squares, with one rather than two square roots for zero.

The quadratic residues form a subgroup of the multiplicative group of a finite field ${\mathbb{F}}_p^{\times }$, with index of a subgroup two. Indeed, one is a square, $(u^2)(v^2)=(uv{)}^2$, and $(u^2{)}^{-1}=(u^{-1}{)}^2$; the inverses exist because the inputs are nonzero in the prime field. Equivalently, the squaring group homomorphism has kernel of a group homomorphism $\{1,-1\}$ and image of a function $Q_p$. The apparent loss of a sign in squaring becomes a precise division of the nonzero arithmetic into two equal parts.

This structure turns a search for square roots into a yes-or-no test: Euler's criterion detects quadratic residues through modular exponentiation, and the Legendre symbol records the result as a sign. For a general integer modulus, the modular congruence $x^2\equiv a(\mathrm{mod}n)$ still makes sense, but the two-root count and the prime field argument need not survive. For example, $x^2\equiv 1(\mathrm{mod}15)$ has four solutions $1,4,11,14$; the Chinese remainder theorem separates the sign choices modulo three and five.

Let $J$ be the plane rotation matrix through $\pi /2$, generating a cyclic group of order four. A quadratic form $a{x}^2+b{y}^2$ has its two coefficients exchanged by a quarter-turn and unchanged by a half-turn. Thus the pullback action on this two-dimensional space gives the displayed group representation: $D(J^k)=D(J{)}^k$, with $D(J{)}^2=I$. It is not a faithful representation, because its kernel of a group homomorphism is $\{I,J^2\}$. Its two invariant coefficient lines correspond to $x^2+y^2$ and $x^2-y^2$, on which a quarter-turn acts by $+1$ and $-1$, respectively. Restricting to this diagonal space is valid for these signed coordinate permutations; a general linear transformation can introduce an $xy$ term.

The orientation-preserving symmetries of a cube are the $3\times 3$ signed permutation matrices with determinant one. There are $3!2^2=24$ of them. Their group action on the four body diagonals identifies this group with the symmetric group $S_4$. Indeed choose diagonal directions $(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)$, whose sum is zero and whose only linear relation has all coefficients equal. A rotation fixing every diagonal sends each direction to itself or its negative; the relation forces all four signs equal. All negative would give $-I$, which has determinant $-1$, so the kernel of a group homomorphism is trivial. Both groups have order 24, giving the identification. Its face permutations have cycle types $1^6$, $1^{2}4$, $1^2{2}^2$, $2^3$ and $3^2$, occurring $1,6,3,6,8$ times respectively.

A finite nonabelian simple group embeds in $A_{n_p}$, where $n_p$ is its number of Sylow subgroups for a prime dividing its order. It is not a p-group by the nontrivial-centre property. Thus its Sylow subgroups are proper and cannot be normal, so $n_p>1$. The conjugation action is nontrivial, and its normal kernel of a group homomorphism must therefore be trivial. The sign of this faithful permutation action must be trivial, since a nontrivial sign homomorphism would inject the group into a group of order two. Lagrange's theorem consequently gives $|G|\mid n_p!/2$.

A subgroup $H\le G$ is a subset of a group $G$ that is itself a group under the same group operation. It must contain the identity element of $G$ and be closed under products and inverse elements. Associativity is inherited from $G$, so it needs no new proof. The whole group and the trivial group $\{e\}$ are always subgroups. A proper subgroup satisfies $H\ne G$.

A useful single-condition test isTo see why it works, choose $a\in H$; then $a{a}^{-1}=e\in H$. Taking $a=e$ gives every inverse element $b^{-1}\in H$. Applying the test to $a$ and $b^{-1}$ then gives $ab\in H$. Conversely, every subgroup satisfies the test by closure under products and inverse elements. This also makes the additive version memorable: a nonempty subset of an additive group is a subgroup exactly when it is closed under differences. See McKernan's MIT lecture 2, Proposition 2.3, for the equivalent closure criterion.

For the square dihedral group $D_4$ above, $R=\{e,r,r^2,r^3\}$ is a cyclic subgroup: products add exponents modulo four, and $(r^k{)}^{-1}=r^{-k}$. The smaller collections $\{e,r^2\}$ and $\{e,s\}$ are subgroups of group order two. But $\{e,r\}$ is not a subgroup, since $r\cdot r=r^2$ lies outside it. A plausible-looking list of symmetries therefore need not constitute a complete collection of reversible operations.

Inside $(\mathbb{Z},+)$, the even numbers $2\mathbb{Z}$ form a subgroup: $2m-2n=2(m-n)$ is an even number. Its two cosets consist of the even numbers and the odd numbers, so its index of a subgroup is two. The parity group homomorphism $k\mapsto k\mathrm{mod}2$ has $2\mathbb{Z}$ as its kernel of a group homomorphism. A subgroup can thus identify exactly which information a composition-preserving map forgets.

Subgroups also describe residual symmetry. In the group action of $D_4$ on the four vertices, the stabilizer subgroup of $(1,1)$ is $\{e,sr\}$: the diagonal reflection $sr$ fixes that vertex, and the other six motions move it. Requiring a feature to stay fixed still leaves a group, because both composition and reversal preserve that requirement. This is why subgroups are useful for studying a group action one feature at a time.