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For a finite group $G$ and a subgroup $H$, Lagrange's theorem statesIn particular $|H|$ divides $|G|$: the left cosets partition $G$, and multiplication by a representative is a bijection from $H$ to each coset. Apply this to the cyclic subgroup $⟨g⟩$. Its size is the order of a group element $g$, so every element order divides $|G|$.

Under the square condition, every group element is its own inverse element. Consequentlyfor all $g,h$. This proves that a group of exponent two is abelian, including the trivial group.

For the fourth-power condition a counterexample to commutativity is the quaternion groupHere $i^2=j^2=k^2=-1$ and $ij=k=-ji$. Thus its six elements outside $\{\pm 1\}$ have order of a group element four, while $1$ and $-1$ have orders one and two. All fourth powers are the identity, yet $ij\ne ji$.

Lagrange's theorem states that for a subgroup $L$ of a finite group $G$, $|G|=[G:L]|L|$. In particular $|L|$ divides $|G|$; the equal-sized cosets partition $G$.

Apply the theorem to $H\cap K$ inside both subgroups. Its order divides both coprime orders, so $H\cap K=\{1\}$. For $h\in H$ and $k\in K$, the group commutator $hk{h}^{-1}{k}^{-1}$ lies in $K$ by $K$ being a normal subgroup, and in $H$ because $k{h}^{-1}{k}^{-1}\in H$. It is therefore the identity. Thus every element of $H$ commutes with every element of $K$.

Define $\theta :H\times K\to G$ by $\theta (h,k)=hk$. The cross-commutation just proved gives $\theta (h,k)\theta (h^{\prime },k^{\prime })=h{h}^{\prime }k{k}^{\prime }=\theta (h{h}^{\prime },k{k}^{\prime })$, so this is a group homomorphism from the direct product of groups. The product assumption makes it surjective. If $hk=1$, then $h=k^{-1}\in H\cap K$, so both entries are the identity and the kernel of a group homomorphism is trivial. HenceThis proves the internal direct product theorem in the present coprime-order setting, rather than merely matching the numbers of elements.

For a nontrivial finite p-group $G$ of order $p^m$ with $m\ge 1$, partition into conjugacy classes. A noncentral class has size $[G:C_G(x)]$, a positive power of $p$ larger than one by Lagrange's theorem. The class equation therefore saysThe center of a group contains the identity, so $|Z(G)|\ge p$, proving its nontriviality. The positive-exponent qualification matters: the one-element group has no nonidentity central element.

If $|G|=p^2$, its center of a group has order $p$ or $p^2$. In the first case, the quotient group $G/Z(G)$ has prime order and is cyclic. Whenever a central quotient group is cyclic, writing all elements as $g^{i}z$ with $z$ central shows that they commute: $(g^{i}z)(g^{j}w)=g^{i+j}zw=(g^{j}w)(g^{i}z)$. Thus that case would already make $G$ Abelian and its center of a group all of $G$, a contradiction. Hence every group of order $p^2$ is Abelian.

If there is an element of order $p^2$, it is a generator of a group for $G$, giving $C_{p^2}$. Otherwise every nonidentity element has order $p$. Pick $a\ne 1$ and $b\notin ⟨a⟩$. Their cyclic subgroups have trivial intersection, and they commute; the $p^2$ distinct products $a^i{b}^j$ exhaust $G$. Hence the classification of groups of order p squared isBoth groups exist and are nonisomorphic, since only the first has an element of order $p^2$.

The conjugacy class of $x$ is its orbit under the conjugation action, and its stabilizer is the centralizer $C_G(x)$. The orbit-stabilizer theorem givesBy Lagrange's theorem, the order of the centralizer divides $p^n=|G|$, so the index is a power of $p$. A conjugacy class has size one exactly when its element belongs to the center of a group.

Separate these singleton classes in the class equation:where the sum takes one representative of each nonsingleton conjugacy class. Every summand is divisible by $p$. Thus $p\mid |Z(G)|$, and the identity ensures $|Z(G)|>0$. Consequently $|Z(G)|\ge p$. This is the nontrivial center of a finite p-group property.

An even permutation in the alternating group $A_6$ has one of the cycle types listed below. In the symmetric group, a type with $m_j$ cycles of length $j$ has centralizer order ${\prod }_j{j}^{m_j}{m}_j!$ and class size $6!/{\prod }_j{j}^{m_j}{m}_j!$: commuting permutations can rotate each cycle and exchange equal-length cycles.

An $S_6$ conjugacy class remains one $A_6$ class if its centralizer contains an odd permutation. Otherwise it splits into two equal classes, since the centralizer is already contained in the index-two subgroup $A_6$. This is the alternating conjugacy class splitting criterion. For the nonsplitting nonidentity types here, odd commuting permutations are respectively a constituent transposition, a transposition of fixed points, the interchange of two 3-cycles, and the constituent 4-cycle. For a 5-cycle with one fixed point, the centralizer is its cyclic group of order five and contains only even permutations.

Thus the conjugacy classes of the alternating group on six letters are:The two 5-cycle representatives lie in different classes. A relabelling that conjugates a 5-cycle to its square acts on its five positions as multiplication by two modulo five, a 4-cycle on the nonzero positions, hence is odd. Every other such conjugator differs by an even centralizer element. The sizes sum to $360=|A_6|$, so the list is exhaustive.

For simplicity of the alternating group on six letters, let $N$ be a normal subgroup. It contains the identity and is a union of full conjugacy classes; also $|N|$ divides $360$ by Lagrange's theorem. A proper subgroup has order at most $180$.

If the class of size 45 is absent, $|N|$ is odd. The odd divisors of $360$ are $1,3,5,9,15,45$. A nontrivial union has order at least 41, but no selection of the even class sizes sums to $44$, so order 45 is impossible. Thus $N=\{1\}$ in this case.

If the size-45 class is present, write $|N|=46+40a+90b+72c$ with $0\le a,c\le 2$ and $0\le b\le 1$. The possible values at most $180$ are $46,86,118,126,136,158,176$, none dividing $360$. Hence no proper nontrivial normal subgroup exists:

Lagrange's theorem says that, for a finite group $G$ and subgroup $H$,Here $[G:H]$ is the number of left cosets. Every $g\in G$ lies in $gH$. If $gH$ and $g^{\prime }H$ meet, write $gh=g^{\prime }{h}^{\prime }$; then $g^{\prime }=gh(h^{\prime }{)}^{-1}$, so $g^{\prime }H=gH$. Hence the left cosets partition $G$. The bijection $h\mapsto gh$ makes each coset have $|H|$ elements. Adding their cardinalities proves Lagrange's theorem.

False. Let $G=S_3$ and $H=K=⟨(12)⟩$. For $g=(13)$, direct multiplication gives the double cosetIt has four elements, and $4\nmid 6=|S_3|$. This is a concrete example of double-coset sizes not dividing group order; Lagrange's theorem applies to subgroups, not to arbitrary double cosets.

The identity belongs to the center of a group. If $z,w$ commute with every $g$, then $zw$ does too; rearranging $gz=zg$ also shows that $z^{-1}$ commutes with $g$. Hence $Z(G)$ is a subgroup. Moreover $gz{g}^{-1}=z$ for $z\in Z(G)$, so it is a normal subgroup.

Suppose $|G|=p^n>1$. The conjugation action partitions $G$ into conjugacy classes. The orbit-stabilizer theorem gives class size $[G:C_G(x)]$, where $C_G(x)$ is the centralizer of $x$. By Lagrange's theorem, this size is a power of $p$; it is one exactly when $x$ is central. Thus the class equation has the formIt follows that $p\mid |Z(G)|$. Since the identity is central, $|Z(G)|\ge 1$, and thereforeThis is the nontrivial-centre property of a finite p-group.

Lagrange's theorem states that, for a finite group $G$ and a subgroup $H$,To prove it, define an equivalence relation by $x\sim y$ when $x^{-1}y\in H$. Its classes are the left cosets $xH$, which therefore partition $G$. Left multiplication by $x$ gives a bijection $H\to xH$, so every coset has $|H|$ elements. Counting this partition gives the formula; in particular $|H|$ divides $|G|$. Applying the theorem to the cyclic subgroup generated by an element shows that its order of a group element divides $|G|$.

Write the given order-two normal subgroup as $N=\{1,z\}$. Conjugation must preserve $N$ and cannot send its nonidentity element to the identity, so $gz{g}^{-1}=z$ for all $g\in G$. Thus $z$ is in the center of a group: a normal subgroup of order two is central.

Choose $g\notin N$. By Lagrange's theorem, its order of a group element is $2$, $p$ or $2p$. It cannot have order two: since $z$ is central, the four distinct elements $1,z,g,zg$ would form a subgroup of order four, whereas $4$ does not divide $2p$ for odd $p$. If $g$ has order $2p$, it already generates $G$. Otherwise $g$ has order $p$, commutes with $z$, and $(gz{)}^k=1$ implies $g^k=z^{-k}\in ⟨g⟩\cap N=\{1\}$. Hence both $p$ and $2$ divide $k$, so $gz$ has order $2p$. In either case

For a group action of a finite group $G$ on a set and a point $x$, the orbit-stabilizer theorem iswhere $G_x=\{g:gx=x\}$ is the stabilizer subgroup. The map $g{G}_x\mapsto gx$ is well defined because elements of the same coset act identically on $x$. If $gx=hx$, then $h^{-1}g\in G_x$, so their cosets agree; every orbit point has the form $gx$, giving surjectivity. This is a bijection. Lagrange's theorem, $|G|=[G:H]|H|$ for any subgroup $H$, yields the counting formula.

Apply the conjugation action of the finite p-group $G$ to its nontrivial normal subgroup $N$. Normality ensures the action stays inside $N$. The singleton orbits of a group action are exactly the elements of $N\cap Z(G)$. Every larger orbit of a group action has size a positive power of $p$, by the orbit-stabilizer theorem and Lagrange's theorem. ThereforeSince $N$ is nontrivial, its order is divisible by $p$. The intersection contains the identity and has cardinality a positive multiple of $p$, so it contains a nonidentity element. This proves the central intersection property of normal subgroups of finite p-groups.

For the proper subgroup $H$, let $H$ act by left multiplication on the left cosets $G/H$. A coset $gH$ is fixed by every $h\in H$ exactly when $g^{-1}Hg\subseteq H$. Both groups have the same finite size, so this inclusion is equality; the fixed cosets are precisely those with $g\in N_G(H)$. Hence their number is $[N_G(H):H]$. All non-singleton orbits of a group action have size divisible by $p$, soSince $H$ is proper, $[G:H]$ is a positive power of $p$ greater than one. The fixed-coset count is nonzero, because $H$ itself is fixed, and is divisible by $p$. Therefore the normalizer strictly contains $H$:This is the normalizer condition for finite p-groups. The proof also covers $H=\{1\}$, whose action fixes all cosets.