Let $P$ and $Q$ be subgroups of orders five and seven, which exist by i. Their normalizers contain $P$ and $Q$, respectively. By Lagrange's theorem, $N_G(P)$ has order $5$ or $35$, and $N_G(Q)$ has order $7$ or $35$. If either has order $35$, its subgroup is normal by iv.

Suppose neither is normal. Then iv gives at least seven distinct subgroups of order five and at least five distinct subgroups of order seven. Distinct subgroups of the same prime order intersect only in the identity, and a subgroup of order five and one of order seven also intersect only in the identity by Lagrange's theorem. Thus $G$ would have at leastnonidentity elements, contradicting $|G|-1=34$. Hence at least one of $P,Q$ is a normal subgroup.

By Lagrange's theorem, every nonidentity group element has order $5$, $7$, or $35$. If some element has order $35$, its seventh and fifth powers have orders $5$ and $7$, respectively, and we are done.

Otherwise every nonidentity element has order $5$ or $7$. Distinct subgroups of a given prime number order intersect only in the identity: a shared nonidentity element generates both. If there were no element of order $5$, the $34$ nonidentity elements would therefore partition into sets of six, the nonidentity elements of the cyclic subgroups of order seven. This would force $6\mid 34$, a contradiction. If there were no element of order $7$, the analogous partition into sets of four would force $4\mid 34$, again a contradiction. Thus elements of both orders exist.

The stabilizer subgroup of $H$ for this conjugation group action is its normalizerIf it is all of $G$, the equality $gH{g}^{-1}=H$ holds for every $g\in G$, which is precisely the definition of a normal subgroup.

If instead $\mathrm{Stab}(H)=H$, the orbit-stabilizer theorem says that the group orbit of $H$ has size $[G:H]$. Every conjugate $gH{g}^{-1}$ is isomorphic to $H$, with group isomorphism $h\mapsto gh{g}^{-1}$, so this group orbit is a subset of $\Omega$. ThereforeThe group action need not be transitive on all of $\Omega$; this one group orbit supplies the lower bound.

For $h\in H$, define $\pi (h)(n)=hn{h}^{-1}$. Since $N$ is a normal subgroup, this sends $N$ into itself. It preserves multiplication, and its inverse is conjugation by $h^{-1}$, so it is an group automorphism of $N$. MoreoverThus $\pi :H\to \mathrm{Aut}(N)$ is a group homomorphism, induced by the group action of conjugation.

If $\pi$ is trivial, every element of $H$ commutes with every element of $N$. Henceis a group homomorphism, since $(n_1{h}_1)(n_2{h}_2)=n_1{n}_2{h}_1{h}_2$. Its image of a group homomorphism is a subgroup containing both $N$ and $H$, and every product $nh$ lies in $⟨N,H⟩$, so its image is exactly $⟨N,H⟩$. If $nh=e$, then $n=h^{-1}\in N\cap H=\{e\}$, so $n=h=e$. Thus $F$ is injective and gives

Write $C_p=⟨g⟩$. A group homomorphism from this cyclic group is determined by the image of $g$, because $\varphi (g^j)=\varphi (g{)}^j$. Each choice $\varphi (g)=g^a$, with $a=0,\dots ,p-1$, gives a well-defined group homomorphism. It is an group automorphism exactly when $g^a$ generates $C_p$.

For a prime number $p$, every $a=1,\dots ,p-1$ is coprime to $p$, so the powers of $g^a$ run through all of $C_p$. The choice $a=0$ gives the trivial group homomorphism and fails to be bijective. Consequently

A map $\varphi :G\to G$ is a group isomorphism if it is a bijection and a group homomorphism, meaningSuch a map from a group to itself is an group automorphism. Its inverse is also a group homomorphism: if $u=\varphi (x)$ and $v=\varphi (y)$, then ${\varphi }^{-1}(uv)=xy={\varphi }^{-1}(u){\varphi }^{-1}(v)$.

Suppose $\sigma$ has order of a group element two. On a cycle of length $k$, ${\sigma }^2$ moves each entry two places. Since ${\sigma }^2=e$, every such $k$ divides two, so $k=1$ or $k=2$. After omitting the fixed points, the disjoint cycle decomposition is therefore a product of disjoint permutation cycles of length two, which are transpositions. There is at least one, because $\sigma \ne e$.

With products acting from right to left, a permutation cycle satisfiesThe right side sends $a_1$ to $a_2$, each $a_j$ to $a_{j+1}$ for $2\le j<k$, and $a_k$ to $a_1$, while fixing all other symbols. This verifies the formula. Applying it to every cycle in the disjoint cycle decomposition expresses any permutation as a product of transpositions. The identity is represented by the empty product.

For each symbol $a$, iterate the permutation $\sigma$. Since the symbol set is finite, some iterate repeats. Applying an inverse power of $\sigma$ shows that the first return is to $a$ itself, givingwith the displayed symbols distinct. This is a permutation cycle. If two such lists share a symbol, applying powers of $\sigma$ shows that they contain exactly the same symbols. Hence these group orbits form a partition of a set $\{1,\dots ,n\}$. On each group orbit, $\sigma$ acts as the corresponding permutation cycle, and these disjoint permutation cycles multiply to $\sigma$.

The disjoint cycle decomposition is unique up to reordering the cycles and cyclically rotating the entries within each cycle, with fixed points optionally omitted. Indeed, each cycle must consist of one of the uniquely determined group orbits, and its successive entries must follow the action of $\sigma$.

For this tripotent linear operator, the relation $T^3=T$ gives $T(T^2-I)=0$. Since $T^2\ne I$, there is $v$ with $(T^2-I)v\ne 0$, and that nonzero vector belongs to $\ker T$. Hence $n_T\ge 1$. Also $T\ne 0$, because $T=0$ would imply $T^2=T$. The rank-nullity theorem in dimension three therefore givesNow suppose $n_T=2$. Then $\mathrm{im}T$ is a line. On this line, $T^2$ is the identity operator: if $y=Tx$, then $T^{2}y=T^{3}x=Tx=y$. Consequently $T$ acts on this line by a scalar $c$ satisfying $c^2=1$. If $c=1$, then $T^{2}x=Tx$ for every $x$, contrary to $T^2\ne T$. Thus $c=-1$.

Choose a nonzero $w\in \mathrm{im}T$, so $Tw=-w$. It does not belong to $\ker T$. A basis $k_1,k_2$ of $\ker T$, followed by $w$, is therefore a basis of $V$, in which

If $T^2=I$, every $v\in V$ has the decompositionwhose first term belongs to $E_1$ and second to $E_{-1}$. Their intersection is zero by ii, so $V=E_1\oplus E_{-1}$ is a direct sum. Choosing bases of the two eigenspaces gives a diagonal matrix with entries $1$ and $-1$.

Since $T^2\ne T$ while $T^2=I$, we have $T\ne I$. Thus $\dim E_{-1}$ is $1$, $2$, or $3$. Reordering the basis gives exactlyFor $-I_3$, $T$ is point inversion through the origin. For $\mathrm{diag}(1,1,-1)$, it fixes the plane $E_1$ pointwise and reverses the complementary line $E_{-1}$: if $v=p+\ell$ in this direct sum, then $Tv=p-\ell$. For $\mathrm{diag}(1,-1,-1)$, it fixes the line $E_1$ pointwise and reverses the complementary plane $E_{-1}$.

These are instances of oblique symmetry of complementary subspaces, allowing oblique eigenspaces. They are a Euclidean reflection in a hyperplane and a rotation in three dimensions through $\pi$ about a line, respectively, when the two eigenspaces are perpendicular. The hypotheses do not require an orthogonal transformation or an orthonormal basis, so that additional metric property cannot be inferred just from the displayed diagonal forms.

The definition of an eigenvalue and the kernel criterion for a singular square matrix giveFor the second equivalence, a linear map from a finite-dimensional vector space to itself has zero kernel exactly when its rank of a linear map is the full dimension, by the rank-nullity theorem, and therefore exactly when it is invertible. This proves both directions.

A real number $\lambda$ is an eigenvalue of $T$ if $Tv=\lambda v$ for some nonzero vector $v\in V$. Its eigenspace isIt is a vector subspace because it contains $0$ and $T(av+bw)=aTv+bTw=\lambda (av+bw)$ whenever $v,w\in E_{\lambda }$ and $a,b\in \mathbb{R}$.

If $v\in E_{\lambda }\cap E_{\mu }$, then $\lambda v=Tv=\mu v$, so $(\lambda -\mu )v=0$. Since $\lambda \ne \mu$, this forces $v=0$. Therefore

Choose an ordered basis $\mathcal{B}$ of $V$ and let $A=[T{]}_{\mathcal{B}}$. Define the characteristic polynomial byIf a different basis gives $B=P^{-1}AP$ for an invertible matrix $P$, multiplicativity of the determinant givesThus the definition is independent of the chosen basis. The alternate convention $\det (A-xI)$ differs by the constant factor $(-1{)}^{\dim V}$ and gives the same roots.

By i and ii,The domain of a function of $S^{\prime }$ is $\mathrm{im}T$. Its rank-nullity theorem gives $r_{S^{\prime }}+n_{S^{\prime }}=\dim \mathrm{im}T=r_T$. Applying the rank-nullity theorem for $T$ now giveswhich is exactly the rank-nullity theorem for the function composition $ST$.

For $u\in U$, the vector $Tu$ already belongs to $\mathrm{im}T$, soTherefore $\ker (ST)=T^{-1}(\ker S^{\prime })$. Since $\ker S^{\prime }=\ker S\cap \mathrm{im}T$, its intersection with $\mathrm{im}T$ is itself. Applying a with $X=\ker S^{\prime }$ yieldsThe inequality follows from the inclusion $\ker S^{\prime }\subseteq \ker S$ and monotonicity of dimension for vector subspaces.

Since $S^{\prime }$ is the restriction of $S$ to $\mathrm{im}T$,This image of a linear map is contained in $\mathrm{im}S$, so $r_{ST}\le r_S$. Also, applying $S$ to a basis of $\mathrm{im}T$ gives a spanning set for $\mathrm{im}{S}^{\prime }$, of size $r_T$. Thus $r_{ST}\le r_T$, and

By i, a free choice of $x_3$ is possible precisely when $\lambda =5$ and $\mu =2$. ThusThe complete solution set of a linear system isThe direction vector spans the kernel, and the displayed straight line is the entire solution set of a linear system.

The elimination in i determines $x_3$ uniquely exactly when $\lambda \ne 5$. The other two coordinates are then also fixed. Explicitly,For $\lambda =5$, the reduced equation is either inconsistent or leaves $x_3$ free, so a unique solution is impossible.

The first two equations give $x_1=1-2x_3$ and $x_2=2-5x_3$. Substitution into the third givesThis equation is inconsistent exactly when its left side is identically zero and its right side is nonzero. Therefore