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Let the lengths in the disjoint cycle decomposition of $\tau$ be ${\ell }_1,\dots ,{\ell }_s$, with fixed points contributing length one. Disjoint cycles commute, and ${\tau }^m=e$ if and only if every cycle raised to $m$ is the identity, equivalently ${\ell }_j$ divides $m$ for every $j$. Hence the order of a group element isIn particular, the identity has order one.

For $n\ge 2$ with $n\ne 6$, bijectivity of $\varphi$ and the previous part implyThe supplied numerical fact forces $k=1$. Thus $\varphi$ permutes the transpositions.

Suppose $n\ge 3$, and write $s_r=\varphi ((1r))$ for $r=2,\dots ,n$. These are distinct transpositions. For $r\ne t$, the product $(1r)(1t)$ has order of a group element three, so $s_r{s}_t$ also has order three. Two distinct transpositions have this property precisely when their two-element sets of moved points meet in one point: disjoint ones have a product of order two, whereas ones sharing one point have a three-cycle as product. The two-element support of a permutation associated with each $s_r$ therefore meets that of every other $s_t$.

We now show that they have a common point. Choose two; the support of a permutation is $\{a,b\}$ for one and $\{a,c\}$ for the other, where $a,b,c$ are distinct. Any further two-element set meeting both either contains $a$ or is $\{b,c\}$. If $\{b,c\}$ occurs, any two-element set meeting all three must itself be one of $\{a,b\},\{a,c\},\{b,c\}$: a set containing $a$ must also contain $b$ or $c$, and a set omitting $a$ must contain both $b,c$. Hence a pairwise-intersecting family with no common point consists of exactly those three edges of a triangle.

For $n\ge 5$, there are $n-1\ge 4$ distinct $s_r$, ruling out that triangle. For $n=3$, the two sets already have a common point. For $n=4$, the triangle alternative would make all three $s_r$ fix the fourth point. They would then generate a proper subgroup of $S_4$, contrary to surjectivity of $\varphi$ and part (a)(iv), which imply that the $s_r$ generate $S_4$. Thus in every case $n\ge 3$ there is a common point $a$, and we may writeThe $b_r$ are distinct and different from $a$, so they exhaust the remaining $n-1$ points. Define a permutation $\sigma$ byThenThe two group homomorphisms $\tau \mapsto \varphi (\tau )$ and $\tau \mapsto {\sigma }^{-1}\tau \sigma$ agree on a generating set of a group, so agree on every product of its elements, hence throughout $S_n$:Finally, $S_1$ is trivial and $S_2$ has a unique nonidentity element, so every automorphism of either is the identity and we can take $\sigma =e$. This completes all degrees $n\ne 6$ and proves that every such automorphism is an inner automorphism.

Assume first $n\ge 2$. All transpositions form a single conjugacy class $T_1$. A group automorphism preserves order of a group element, so the image of a transposition is a nonidentity involution. By the order formula, its disjoint cycle decomposition contains only one-cycles and two-cycles, with at least one two-cycle. Thus it lies in some $T_k$, where $1\le k\le \lfloor n/2\rfloor$.

Moreover an automorphism maps the entire conjugacy class onto the conjugacy class of the image: for a fixed transposition $t$,and $\varphi (g)$ runs over all of $S_n$ by surjectivity. Part (a)(i) identifies that latter conjugacy class as $T_k$. Hence $\boxed{\varphi (T_1)=T_k}$, with equality rather than just inclusion. If $n=1$, $T_1$ is empty and the equality holds trivially with $k=1$.

A group has a closed binary operation, associativity, an identity, and an inverse for each element. For a finite group, Lagrange's theorem states $|G|=[G:H]|H|$ for each subgroup $H$, so $|H|$ divides $|G|$.

The order of a group element $g$ is the least positive $n$ with $g^n=e$, or infinity if none exists. In a finite group the powers form the cyclic subgroup $⟨g⟩$ of that order; Lagrange's theorem proves $\mathrm{ord}(g)\mid |G|$. If $|G|=p$ is prime and $g\ne e$, its order must be $p$. Thus $g$ generates the whole group, proving $G\cong C_p$.

If every nonidentity element has order two, every element is its own inverse. Then $ab=(ab{)}^{-1}=b^{-1}{a}^{-1}=ba$. This proves that a group of exponent two is abelian.

The order of a finite group $G$ is its number of elements. Lagrange's theorem states that for a subgroup $K\le G$,so $|K|$ divides $|G|$. In particular the order of a group element divides $|G|$, because it is the order of its generated cyclic subgroup.

For $|G|=4$, if some element has order four it generates the whole group, which is therefore the cyclic group $C_4$. Otherwise every nonidentity element has order two. Choose two distinct ones $a,b$ and put $c=ab$. Cancellation shows $c\notin \{e,a,b\}$, so $c$ is the remaining element. Since $c^{-1}=c$,The other products follow: $ac=ca=b$, $bc=cb=a$, and $a^2=b^2=c^2=e$. Thus the second possibility is the Klein four-group. The two groups are not isomorphic because only $C_4$ has an element of order four. This proves the order-four case of the classification of groups of order p squared.

A group homomorphism $\Phi :G\to H$ obeys $\Phi (gh)=\Phi (g)\Phi (h)$. It maps identity to identity and inverses to inverses. Its group kernel $K=\{g:\Phi (g)=e_H\}$ contains $e_G$ and is closed under $a{b}^{-1}$, so the subgroup test makes it a subgroup. For $k\in K$ and $g\in G$,Thus $gK{g}^{-1}\subseteq K$; applying the same argument to $g^{-1}$ gives equality. Hence $K$ is a normal subgroup.

Define $\overline{\Phi }:G/K\to H$ by $\overline{\Phi }(gK)=\Phi (g)$. If $gK=hK$, then $h^{-1}g\in K$, giving $\Phi (g)=\Phi (h)$, so this is well-defined. Coset multiplication shows it is a group homomorphism. Its kernel is just the identity coset, since $\Phi (g)=e_H$ means $g\in K$. It is therefore injective, and the assumed surjectivity of $\Phi$ makes it surjective as well. This proves the first isomorphism theorem here:

In the quaternion group, the element $-1=i^2=j^2=k^2$ commutes with each of $i,j,k$, because an element commutes with its square. It then commutes with their negatives too, and with $\pm 1$. The set $N=\{1,-1\}$ is a subgroup because $(-1{)}^2=1$, and it is central, hence a normal subgroup.

The four quotient cosets are $N,iN,jN,kN$. They square to $N$, and $(iN)(jN)=kN$. Reversed products have the same coset: for example $(ij{)}^{-1}=k^{-1}=-k$, while $j^{-1}{i}^{-1}=(-j)(-i)=ji$, so $ji=-k$. The same calculation gives $kj=-i$ and $ik=-j$. After signs are identified, these are precisely the products in the Klein four-group. DefineThis preserves all products, is surjective, and has kernel exactly $N$. The first isomorphism theorem therefore givesThis is the quaternion group modulo its centre: the reverse-product relations also show that none of $\pm i,\pm j,\pm k$ is central, so $Z(Q_8)=N$.

The group order of a finite group is its number of elements, $|G|$. The order of a group element $g$ is the least positive integer $r$ with $g^r=e$. Such an $r$ exists in a finite group because two powers must coincide, and cancellation then gives a positive power equal to the identity element.

For the direct product of groups, the componentwise operation is closed because each factor is closed. Associativity holds in each coordinate:The identity element is $(e_1,e_2)$, and the inverse element of $(g,h)$ is $(g^{-1},h^{-1})$. Thus all group axioms hold. For finite factors, the multiplication principle gives

For ${\mathbb{Z}}_2\times {\mathbb{Z}}_2$, write $E=(e,e)$, $u=(w,e)$, $v=(e,w)$, $z=(w,w)$, with $w^2=e$. Its Cayley table, with the row element multiplying the column element, isThis is the Klein four-group: every nonidentity element has order two.

In the cyclic group ${\mathbb{Z}}_4=⟨a⟩$, the set $\{e,a^2\}$ is closed because $(a^2{)}^2=e$, and both elements have inverses in the set. It is therefore a proper subgroup of order two. By Lagrange's theorem, a proper subgroup has order one or two. An order-two subgroup must be generated by an element of order two. Neither $a$ nor $a^3$ has order two, whereas $a^2$ does. HenceThe PDF's claim needs “nontrivial”: $\{e,a^2\}$ is the only nontrivial proper subgroup, but $\{e\}$ is also a proper subgroup under the usual definition.

A group isomorphism preserves element orders: $\varphi (g{)}^r=e$ if and only if $g^r=e$, using the homomorphism property and injectivity. Since ${\mathbb{Z}}_4$ has elements of order four and the Klein four-group does not,

The order of a group element $g$ is the least positive integer $q$ such that $g^q=e$, or infinity if there is no such integer. In a finite group such a $q$ exists, because two powers among $e,g,g^2,\dots$ must coincide.

The powers $e,g,\dots ,g^{q-1}$ are distinct: a repetition with exponents differing by less than $q$ would contradict its minimality. Division of any exponent by $q$ shows that these are all the elements of the cyclic subgroup $⟨g⟩$. Thus $|⟨g⟩|=q$. Applying Lagrange's theorem givesIn particular the identity has order one, and every nonidentity element has order greater than one. The divisibility assertion here concerns finite group order.

Write the cyclic factors additively. The order of a group element $(r,s)$ in a direct product of groups is the least common multiple of its component orders, since a power is the identity exactly when it is the identity in both coordinates.

In $C_2\times C_3$, $(1,1)$ has order $\mathrm{lcm}(2,3)=6$, so it generates all six pairs. An explicit group isomorphism isIt is well defined and multiplicative in additive notation; its kernel is zero because an integer divisible by both two and three is divisible by six. Equal finite orders then give bijectivity.

In $C_2\times C_4$, every element has order dividing four. A generator of $C_8$ has order eight, and isomorphisms preserve element orders, soFor positive integers $n,m$, the maximum element order in $C_n\times C_m$ is $\mathrm{lcm}(n,m)=nm/\gcd (n,m)$, attained by a pair of generators. The product is cyclic of order $nm$ exactly when that number is $nm$. Thus cyclicity of a product of two finite cyclic groups givesFor coprime factors the map $[k{]}_{nm}\mapsto ([k{]}_n,[k{]}_m)$ gives the isomorphism, the group form of the Chinese remainder theorem.

For every positive integer $m$, the group laws give $(g^{-1}{)}^m=(g^m{)}^{-1}$. ConsequentlyThe sets of positive exponents returning the two elements to the identity element are identical. If this set is nonempty, its least member is the order of a group element in both cases. If it is empty, neither element has finite order. Therefore, including the infinite-order case,For a direct divisibility argument in the finite case, if $g$ has order $n$, then $(g^{-1}{)}^n=e$, so its inverse’s order divides $n$. Reversing the roles shows $n$ divides that order, forcing equality. This is the general fact that an element and its inverse have the same order; it also reflects $⟨g⟩=⟨g^{-1}⟩$ as cyclic subgroups.

Let $H$ be a proper subgroup. By Lagrange's theorem, $|H|$ divides $21=3\cdot 7$. Since $H\ne G$, its possibilities areThe subgroup of order one is generated by the identity and is cyclic. For either prime order $p=3$ or $7$, choose $h\in H$ other than the identity. Its order of a group element divides $p$ and is not one, so it is $p$. Thus its powers exhaust $H$, proving $H=⟨h⟩$.

ThereforeThis uses the fact that a group of prime order is cyclic. It does not require the order-21 group itself to be Abelian or cyclic.

A group is a set $G$ equipped with a binary operation satisfying the group axioms: closure; associativity, $(gh)k=g(hk)$; an identity element $e$ with $eg=ge=g$; and an inverse element $g^{-1}$ for every $g$, with $g{g}^{-1}=g^{-1}g=e$.

A subgroup $H$ is a subset that is a group under the restricted operation. Equivalently, it is nonempty and contains $x{y}^{-1}$ whenever $x,y\in H$. A generating set of a group is a set whose elements and inverses can be multiplied in finite words to produce every group element.

The group order is its number of elements, $|G|$. The order of a group element $g$ is the least positive integer $r$ with $g^r=e$, or infinity if there is none. Lagrange's theorem states that for a finite group and subgroup $H$, $|H|$ divides $|G|$ and $|G|=[G:H]|H|$. In particular, every finite element order divides the group order.

A cyclic group is generated by one element, so every element is a power of that generator. A group isomorphism is a bijective group homomorphism: a bijection $\varphi :G\to G^{\prime }$ satisfying $\varphi (gh)=\varphi (g)\varphi (h)$ for all $g,h$.

By Lagrange's theorem, the possible values of the order of a group element areThese are possibilities across groups of order eight; they need not all occur in one group.

If every element has order at most two, every element equals its own inverse. Thus for any $x,y$,so $G$ is an abelian group. Choose $a\ne e$, and then $b\notin ⟨a⟩$. The four elements $e,a,b,ab$ are distinct and form a subgroup $H$: commutativity and $a^2=b^2=e$ prove closure and inverses. Choose $c\notin H$. The elements of $H$ and $Hc$ are disjoint, since $hc\in H$ would imply $c\in H$. These two four-element sets therefore exhaust $G$.

Every element consequently has a unique expression $a^i{b}^j{c}^k$, with $i,j,k\in \{0,1\}$. The map from triples to these products is bijective and respects multiplication by addition of exponents modulo two. Hencea direct product of groups and an elementary abelian group.

Each reflection has order two, giving exactly the three order-two subgroupsConjugation by $r$ sends $r^{j}s$ to $r^{j+2}s$, cycling the three reflections. None of these subgroups is therefore normal.

The two nonidentity rotations generate the same order-three subgroup:It is invariant under conjugation by $r$, and $srs=r^{-1}$ shows invariance under $s$ as well; since $r,s$ generate the group, this proves normality. The complete partition into conjugacy classes isThe rotations are mutually conjugate by $s$, and the reflections by $r$; no rotation is conjugate to a reflection because conjugation preserves the order of a group element.

Write $G=\{e,a,b,c\}$. By Lagrange's theorem, the order of a group element divides four. Since no element has order four, all three nonidentity elements have order two. Thus every element is its own inverse. For any $x,y\in G$,so the group is Abelian. Also $ab$ cannot equal $e,a$ or $b$, by cancellation and the fact that $a,b$ are distinct nonidentity elements. Hence $ab=c$. The remaining products are forced, givingThe map $e\mapsto (0,0)$, $a\mapsto (1,0)$, $b\mapsto (0,1)$, $c\mapsto (1,1)$ preserves this table under componentwise addition modulo two. Thereforethe Klein four-group, also called the Vierergruppe.

The group is the direct product of groups ${\mathbb{Z}}_p\times {\mathbb{Z}}_q$. Repeatedly adding $(1,1)$ returns to $(0,0)$ for the first time when both $p$ and $q$ divide the step number. Its order of a group element is therefore $\mathrm{lcm}(p,q)$. If $p,q$ are coprime, this is $pq$, so it generates every element:Explicitly, $[t{]}_{pq}\mapsto ([t{]}_p,[t{]}_q)$ is the Chinese remainder theorem group isomorphism.

If $d=\gcd (p,q)>1$, every pair satisfiesThus every element order divides $pq/d<pq$, and no element can generate the whole group. HenceThis is the cyclicity of a product of two finite cyclic groups.

The group order is its cardinality, $\boxed{|G|=\#\{g:g\in G\}}$. The order of a group element $g$ iswhere $e$ is the identity element. In a finite group such an $n$ always exists, since two nonnegative powers of $g$ must coincide.

Choose $g\ne e$ in a group of prime group order $p$. Its order of a group element divides $p$ by (c) and is not $1$, hence equals $p$. Consequently its cyclic subgroup already contains all elements of $G$:Thus every nonidentity element is a group generator.

In additive notation every element $(a,b)\in {\mathbb{Z}}_2\times {\mathbb{Z}}_4$ obeys $4(a,b)=(0,0)$. Its order of a group element therefore divides four, whereas $1\in {\mathbb{Z}}_8$ has order eight. A group isomorphism preserves element orders, soFor the second product, define the group homomorphismIt is well-defined because multiples of six are divisible by both two and three. Its kernel contains only zero: simultaneous divisibility by two and three implies divisibility by six. Thus $\Psi$ is injective, and equal finite group orders make it surjective. Explicitly the six images areThereforeFor general positive integers $n,m$, the order of a direct-product element is the least common multiple of its component orders. Hence every element order divides $\mathrm{lcm}(n,m)=nm/\gcd (n,m)$, while $(1,1)$ attains this order. It generates the whole product precisely when $n,m$ are coprime:This is the cyclicity of a product of two finite cyclic groups, with the explicit isomorphism furnished by the Chinese remainder theorem.

Write $C_n=⟨g⟩$ and set $N=⟨g^p⟩$. The order of $g^p$ is $n/\gcd (n,p)=m$, so $N\cong C_m$. Every subgroup of an abelian group is a normal subgroup. The quotient group $C_n/N$ is generated by $gN$ and has order $n/m=p$, hence is $C_p$ and is a simple group. By the subgroup correspondence theorem, there is no proper normal subgroup strictly between $N$ and $C_n$. Therefore $N$ is a maximal normal subgroup with the required quotient group.

The exact orders printed in the question are inconsistent. The source PDF says that the orders of group elements $a,b,c$ are $4,12,24$, respectively, and also imposes $3a+2b=0$. Multiplying this relation by $6$ and using $12b=0$ gives $18a=0$. Since $4a=0$ also holds, subtracting $4(4a)=0$ gives $2a=0$, contradicting the stated order of a group element $4$. Thus

The natural abelian group presentation interpretation is to impose the relations $4a=12b=24c=0$ and $3a+2b=0$, with no further relations, without requiring the orders of the images of the generators to stay $4,12,24$. Under this interpretation the calculation is as follows. Since $3a=-a$ follows from $4a=0$, the mixed relation gives $a=2b$. Consequently $8b=0$, and together with $12b=0$ this gives $4b=0$. Conversely, $a=2b$, $4b=0$, and $24c=0$ imply every original relation. Eliminating $a$ leaves two independent cyclic groups:To verify that these relations force no further identifications, map $a,b,c$ to $(2,0),(1,0),(0,1)$ in ${\mathbb{Z}}_4\oplus {\mathbb{Z}}_{24}$. All defining relations hold, and the inverse group homomorphism sends the two standard generators to $b,c$. These group homomorphisms are mutually inverse. The divisibility condition $4\mid 24$ gives the requested invariant factors of a finitely generated module over $\mathbb{Z}$. In this presented group, $a,b,c$ actually have orders $2,4,24$.

Equivalently, the relation matrixhas nonzero Smith normal form entries $1,4,24$, with the trivial ${\mathbb{Z}}_1$ factor omitted. The displayed direct sum of modules is the answer to the presentation interpretation; the inconsistency of the literal exact-order formulation remains.