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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:

True. The sign of a permutation is a group homomorphism $\mathrm{sgn}:S_3\to \{1,-1\}$, as follows, for example, from the multiplicative determinant of permutation matrices. Its kernel of a group homomorphism is the alternating groupA transposition maps to $-1$, so this homomorphism is onto.

A permutation representation is a group homomorphism $\rho :G\to \mathrm{Sym}(\Omega )$, equivalently a group action on a set $\Omega$. The left regular action takes $\Omega =G$ and $\rho (g)(h)=gh$. Each such map is a bijection, and $\rho (g_1{g}_2)=\rho (g_1)\rho (g_2)$. If $\rho (g)$ is the identity, evaluating at the identity element gives $g=1$. Thus the representation is faithful. Numbering the elements of $G$ identifies $\mathrm{Sym}(G)$ with the symmetric group $S_n$, proving Cayley theorem:For a nonabelian simple group, compose this faithful representation with the sign of a permutation homomorphism $S_n\to \{1,-1\}$. Its kernel is a normal subgroup of $G$, hence either trivial or all of $G$. A trivial kernel would embed $G$ in a group of order two, contradicting noncommutativity. Therefore every represented element is even, and the same representation embeds $G$ in the alternating group $A_n$.

Finally let $S_3$ act on the two cosets of its normal subgroup $A_3$. An even permutation fixes both cosets, whereas an odd permutation exchanges them. This gives $S_3\to S_2$ with kernel $A_3$; explicitly, send even permutations to the identity and odd permutations to $(12)$.

$\boxed{A_5\text{ is not a nontrivial free product}.}$ If $g\ne 1$ and $h\ne 1$ belong to different factors, the powers $(gh{)}^r$ have distinct normal forms of length $2r$. Every nontrivial free product is therefore an infinite group, whereas the alternating group $A_5$ is a finite group of order $60$.

With rightmost-first composition, a permutation cycle satisfiesTracing each $a_j$ verifies the equality, and every other letter is fixed. Thus every cycle is a product of transpositions; decomposing a permutation into disjoint permutation cycles proves that every element of $S_n$ is such a product. The identity is the empty product.

To establish that parity is well defined, use the permutation matrix $P_{\sigma }$ defined by $P_{\sigma }{e}_i=e_{\sigma (i)}$. Its determinant is $\pm 1$, and $P_{\sigma \tau }=P_{\sigma }{P}_{\tau }$. A transposition swaps two columns of the identity matrix and has determinant $-1$. Therefore if $\sigma$ is expressed as a product of $r$ transpositions,The left side depends only on $\sigma$, so any two decompositions have the same parity. DefineMultiplicativity of the determinant makes this the sign homomorphism. Its kernel of a group homomorphism is the alternating groupIt is a normal subgroup consisting of the even permutations. For $n\ge 2$ the sign map is surjective because a transposition has sign $-1$, so $[S_n:A_n]=2$. For $n=1$ both groups are trivial.

A finite nonabelian simple group cannot be a p-group: a nontrivial finite p-group has nontrivial centre of a group, since its class equation makes every noncentral conjugacy class size divisible by $p$, and hence makes the centre size a positive multiple of $p$. That centre is a normal subgroup, so must be the whole group if $G$ is a simple group, making it abelian; an abelian simple group has prime order. Therefore every Sylow subgroup here is nontrivial and proper. It cannot be normal, so $n_p>1$.

To obtain a simple group embedding from Sylow conjugation, the conjugation action on Sylow subgroups gives a group homomorphism $\rho :G\to S_{n_p}$. This group action is transitive by the Sylow theorems and is nontrivial since $n_p>1$. Its kernel of a group homomorphism is a normal subgroup, so the defining property of a simple group makes the kernel of a group homomorphism trivial: $\rho$ is an embedding. Now compose with the sign of a permutation $S_{n_p}\to \{\pm 1\}$. A nontrivial composite would again be injective because $G$ is a simple group, embedding $G$ in a group of order two, impossible for a nonabelian group. Thus $\rho (G)$ lies in the alternating group $A_{n_p}$. By Lagrange's theorem,The argument only invokes the factorial formula once $n_p\ge 2$; the impossible case $n_p=2$ itself is also excluded by the embedding.

If a monic polynomial $f$ has roots ${\alpha }_1,\dots ,{\alpha }_n$ in a splitting field, its polynomial discriminant isIt is a symmetric polynomial in the roots, so it belongs to the base field; it is nonzero exactly when $f$ is a separable polynomial.

Putthe Vandermonde determinant. A root permutation $\sigma$ sends $D$ to $\mathrm{sgn}(\sigma )D$. Hence the Galois group of a polynomial is contained in the alternating group $A_n$ exactly when every Galois automorphism fixes $D$, which by the fixed-field property is equivalent to $D\in K$. This implies that $\mathrm{disc}(f)=D^2$ is a square in $K$. Conversely, if $\mathrm{disc}(f)=d^2$ for $d\in K$, then $D^2=d^2$, so $D=\pm d\in K$. The Galois group therefore fixes $D$ and consists of even permutations. This proves the discriminant criterion for an alternating Galois group.

Now letThe rational root theorem shows that none of $\pm 1,\pm 2$ is a root, so this cubic is irreducible over $\mathbb{Q}$. The discriminant of a depressed cubic isThis is not a square in $\mathbb{Q}$, so the Galois group of an irreducible cubic is $S_3$ over $\mathbb{Q}$.

Over $K=\mathbb{Q}(\sqrt{-19})$, the discriminant is the squareThe cubic remains irreducible: a root in the degree-two extension $K/\mathbb{Q}$ would generate over $\mathbb{Q}$ both a degree-three field, by irreducibility, and a subfield of a degree-two field, contradicting the tower law for field extensions. Its Galois group over $K$ is consequently $A_3\cong C_3$.

The sign homomorphism sends an even permutation to $1$ and an odd permutation to $-1$; its kernel is the alternating group.

A symmetric group is the group of all bijections from a set to itself, with function composition as its operation.

A regular tetrahedron has a proper rotational symmetry group $T$ of order 12, isomorphic to the alternating group $A_4$ acting on its four vertices. Its full spatial symmetry group $T_d$ has order 24 and is isomorphic to the symmetric group $S_4$; the additional operations reverse orientation. Three-fold vertex/face-axis rotations and half-turns about opposite-edge midpoints generate the proper group. The distinction matters when a field symmetry uses only proper rotations, but a scalar density also admits reflections.