Tag Archives: symmetric group

Compute in a group ring

Consider the following elements of the group ring $\mathbb{Z}/(3)[S_3]$: $\alpha = 1(2\ 3) + 2(1\ 2\ 3)$ and $\beta = 2(2\ 3) + 2(1\ 3\ 2)$. Compute $\alpha + \beta$, $2\alpha - 3\beta$, $\alpha\beta$, $\beta\alpha$, and $\alpha^2$.

Evidently,

1. $\alpha + \beta = 2(1\ 2\ 3) + 2(1\ 3\ 2)$
2. $2\alpha - 3\beta = 2\alpha = 2(2\ 3) + (1\ 2\ 3)$
3. $\alpha\beta = 2(1\ 2) + 1(1\ 2\ 3)$
4. $\beta\alpha = 1(1\ 3) + 2(1\ 3\ 2)$
5. $\alpha^2 = 1(1) + 2(1\ 2) + 2(1\ 3) + 1(1\ 3\ 2)$

Compute in a group ring

Consider the following elements of the integral group ring $\mathbb{Z}[S_3]$: $\alpha = 3(1\ 2) - 5(2\ 3) + 14(1\ 2\ 3)$ and $\beta = 6(1) + 2(2\ 3) - 7(1\ 3\ 2)$. Compute the following elements: $\alpha + \beta$, $2\alpha - 3\beta$, $\alpha\beta$, $\beta\alpha$, and $\alpha^2$.

Evidently,

1. $\alpha + \beta = 6(1) + 3(1\ 2) - 3(2\ 3) + 14(1\ 2\ 3) - 7(1\ 3\ 2)$
2. $2\alpha - 3\beta = -18(1) + 6(1\ 2) - 16(2\ 3) + 28(1\ 2\ 3) + 21(1\ 3\ 2)$
3. $\alpha\beta = -108(1) + 81(1\ 2) - 21(1\ 3) - 30(2\ 3) + 90(1\ 2\ 3)$
4. $\beta\alpha = -108(1) + 18(1\ 2) + 63(1\ 3) - 51(2\ 3) + 84(1\ 2\ 3) + 6(1\ 3\ 2)$
5. $\alpha^2 = 34(1) - 70(1\ 2) - 28(1\ 3) + 42(2\ 3) - 15(1\ 2\ 3) + 181(1\ 3\ 2)$

An automorphism of Sym(6) that is not inner

This exercise exhibits an automorphism of $S_6$ that is not inner (hence it shows that $[\mathsf{Aut}(S_6) : \mathsf{Inn}(S_6)] = 2$. Let $u_1 = (1\ 2)(3\ 4)(5\ 6)$, $u_2 = (1\ 4)(2\ 5)(3\ 6)$, $u_3 = (1\ 3)(2\ 4)(5\ 6)$, $u_4 = (1\ 2)(3\ 6)(4\ 5)$, and $u_5 = (1\ 4)(2\ 3)(5\ 6)$. Show that $u_1, \ldots, u_5$ satisfy the following relations:

• $u_i^2 = 1$ for all $i$,
• $(u_iu_j)^2 = 1$ for all $i,j$ with $|i-j| \geq 2$, and
• $(u_iu_{i+1})^3 = 1$ for all $i \in \{1,2,3,4\}$.

Deduce that $S_6 = \langle u_1, \ldots, u_5 \rangle$ and that the map $(1\ 2) \mapsto u_1$, $(2\ 3) \mapsto u_2$, $(3\ 4) \mapsto u_3$, $(4\ 5) \mapsto u_4$, $(5\ 6) \mapsto u_5$ extends to an automorphism of $S_6$ (which is clearly not inner since it does not preserve cycle shape.

Showing that $u_i$ satisfy the given relations is a straightforward computation. Moreover, they generate $S_6$ since $u_3u_5u_1 = (6\ 5)$ and $u_1u_2u_4 = (6\ 5\ 2\ 3\ 4\ 1)$.

The automorphism group of the quaterion group is isomorphic to Sym(4)

Prove that $\mathsf{Aut}(Q_8) \cong S_4$.

We already know that $|\mathsf{Aut}(Q_8)| = 24$ by counting the possible images of, say, $i$ and $j$. By a previous theorem, it suffices to show that no automorphism of $Q_8$ has order 6.

Let $\varphi$ be an automorphism of $Q_8$. Now $Q_8$ has 6 elements of order 4, one element of order 2, and one identity; $\varphi$ must fix the identity and the element of order 2, so that we can consider $\varphi$ as an element of $S_A$, where $A = \{i,j,k,-i,-j,-k\}$. Suppose now that some automorphism $\varphi$ has order 6. Now $\varphi(i) = a$, where $a \notin \{i,-i\}$, and $\varphi(a) = b$, where $b \notin \{i,-i,a,-a\}$. Clearly $\varphi(-i) = -a$ and $\varphi(-a) = -b$. If $\varphi(b) = i$, then $\varphi = (i\ a\ b)(-i\ -a\ -b)$ and $\varphi$ has order 3; thus $\varphi(b) = -i$, and we have $\varphi = (i\ a\ b\ -i\ -a\ -b)$. Now $\varphi^3 = (i\ -i)(a\ -a)(b\ -b)$. Note that $ia \in \{b,-b\}$. If $ia = b$, then $ia = (-i)(-a)$ $= \varphi^3(i)\varphi^3(a)$ $= \varphi^3(ia)$ $= \varphi^3(b) = -b$. Similarly, if $ia = -b$, then $ia = b$. Both cases yield a contradiction, so that no automorphism of order 6 exists.

In addition, we record this handy lemma.

Lemma: Let $G$ be a group and let $K \leq G$ be characteristic. Then $\mathsf{Aut}(G)$ acts on $G/K$ by $\varphi \cdot xK = \varphi(x)K$. Proof: First we need to show well-definedness: if $xK = yK$, we have $xy^{-1} \in K$. Then $\varphi(xy^{-1}) = \varphi(x)\varphi(y)^{-1} \in K$ since $K$ is characteristic in $G$, so that $\varphi(x)K = \varphi(y)K$. Moreover, $1 \cdot xK = 1(x)K = xK$ and $(\varphi \circ \psi) \cdot xK = (\varphi \circ \psi)(x)K = \varphi(\psi(x))K$ $= \varphi \cdot (\psi(x)K)$ $= \varphi \cdot (\psi \cdot xK)$. $\square$

Exhibit a presentation for Sym(4) with two generators

Establish a finite presentation for $S_4$ with 2 generators.

Recall that $S_4$ is generated by $\alpha = (1\ 2)$ and $\beta = (1\ 2\ 3\ 4)$. Evidently, these elements satisfy the relations $\alpha^2 = \beta^4 = (\alpha\beta)^3 = 1$. We will now show that any group presented by $\langle a, b \ |\ a^2 = b^4 = (ab)^3 \rangle$ has at most 24 elements.

• There is one reduced word of length 0: 1.
• There are two reduced words of length 1: $a$ and $b$.
• There are at most four reduced words of length 2: $a^2 = 1$, $ab$, $ba$, and $b^2$. We see that at most three of these are reduced.
• There are at most six reduced words of length 3: $aba$, $ab^2$, $ba^2 = b$, $bab$, $b^2a$, $b^3$. We see that at most five of these are reduced.
• There are at most ten reduced words of length 4: $aba^2 = ab$, $abab$, $ab^2a$, $ab^3$, $baba = ab^3$, $bab^2$, $b^2a^2 = b^2$, $b^2ab$, $b^3a = abab$, and $b^4 = 1$. At most five of these are reduced.
• There are at most ten reduced words of length 5: $ababa = b^3$, $abab^2$, $ab^2a^2 = ab^2$, $ab^2ab$, $ab^3a = bab$, $ab^4 = a$, $bab^2a$, $bab^3$, $b^2aba = bab^3$, and $b^2ab^2$. We see that at most five of these are reduced.
• There are at most ten reduced words of length 6: $abab^2a$, $abab^3$, $ab^2aba = abab^3$, $ab^2ab^2$, $bab^2a^2 = bab^2$, $bab^2ab = ab^2a$, $bab^3a = b^2ab$, $bab^4 = ba$, $b^2ab^2a = ab^2ab^2$, and $b^2ab^3 = abab^2a$. We see that at most three of these are reduced.
• There are at most six reduced words of length 7: $abab^2a^2 = abab^2$, $abab^2ab = b^2a$, $abab^3a = ab^2ab$, $abab^4 = aba$, $ab^2ab^2a = b^2ab^2$, and $ab^2ab^3 = ab^2ab$. We see that none of these is reduced.

Thus any group with this presentation has at most 24 elements, and in fact we have $S_4 = \langle a,b \ |\ a^2 = b^4 = (ab)^3 = 1 \rangle$.

A group of order 24 with no elements of order 6 is isomorphic to Sym(4)

Show that a group of order 24 with no element of order 6 is isomorphic to $S_4$.

Note that $24 = 2^3 \cdot 3$, so that Sylow’s Theorem forces $n_2(G) \in \{1,3\}$ and $n_3(G) \in \{1,4\}$.

Suppose $n_3(G) = 1$. If $n_2(G) = 1$, then by the recognition theorem for direct products, $G \cong P_2 \times P_3$, where $P_2$ and $P_3$ are Sylow 2- and 3-subgroups of $G$, respectively. By Cauchy, there exist elements $x \in P_2$ and $y \in P_3$ of order 2 and 3, so that $xy$ has order 6, a contradiction. Suppose now that $n_2(G) = 3$. Since $n_2(G) \not\equiv 1$ mod 4, there exist $P_2,Q_2 \in \mathsf{Syl}_2(G)$ such that $P_2 \cap Q_2$ is nontrivial and its normalizer has order $2^3 \cdot 3$, by a previous theorem. Thus $P_2 \cap Q_2 \leq G$ is normal. Note that $|P_2 \cap Q_2|$ is either 2 or 4.

Suppose $|P_2 \cap Q_2| = 4$. Then at most $7 + 3 + 7$ nonidentity elements of $G$ are contained in Sylow 2-subgroups, and 3 elements are contained in Sylow 3-subgroups. This leaves 4 elements not of prime power order, one of which must have order 6- a contradiction.

Suppose now that $|P_2 \cap Q_2| = 2$. By the N/C Theorem, $G/C_G(P_2 \cap Q_2) \leq \mathsf{Aut}(P_2 \cap Q_2) \cong 1$, so that $P_2 \cap Q_2$ is central in $G$. By Cauchy, there exist elements $x \in P_2 \cap Q_2$ and $y \in G$ of order 2 and 3, so that $xy$ has order 6, a contradiction.

Thus we may assume $n_3(G) = 4$. Let $P_3 \leq G$ be a Sylow 3-subgroup and let $N = N_G(P_3)$. The action of $G$ on $G/N$ yields a permutation representation $G \rightarrow S_4$ whose kernel $K$ is contained in $N$. Recall that normalizers of Sylow subgroups are self normalizing, so that $N$ is not normal in $G$. Moreover, we have $|N| = 6$. We know from the classification of groups of order 6 that $N$ is isomorphic to either $Z_6$ or $D_6$; however, in the first case we have an element of order 6, a contradiction. Thus $N \cong D_6$. We know also that the normal subgroups of $D_6$ have order 1, 3, or 6. If $|K| = 6$, then $K = N$ is normal in $G$, a contradiction. If $|K| = 3$, then by the N/C theorem we have $G/C_G(K) \leq \mathsf{Aut}(Z_3) \cong Z_2$. In particular, $C_G(K)$ contains an element of order 2, so that $G$ contains an element of order 6, a contradiction.

Thus $K = 1$, and in fact $G \leq S_4$. Since $|G| = |S_4| = 24$ is finite, $G \cong S_4$.

Computation of some Frattini subgroups

Compute $\Phi(S_3)$, $\Phi(A_4)$, $\Phi(S_4)$, $\Phi(A_5)$, and $\Phi(S_5)$.

We begin with some lemmas.

Lemma 1: If $H \leq S_4$ is a subgroup of order 12, then $H = A_4$. Proof: By Cauchy, $H$ contains an element of order 2 and an element of order 3. If some element of order 2 is a product of two 2-cycles, then $A_4 \leq H$, and by counting elements we have $H = A_4$. Suppose no element in $H$ is a product of two 2-cycles; then also no element of $H$ is a 4-cycle, since the square of a 4-cycle is a product of two 2-cycles. Suppose now that $H$ contains two 3-cycles which are not inverses of one another; from the subgroup lattice of $A_4$, then, $A_4 \leq H$; since $H$ also contains a 2-cycle, we have $H = S_4$, a contradiction. Suppose then that $H$ contains exactly two 3-cycles, $(a\ b\ c)$ and $(a\ c\ b)$. then $H$ must contain nine 2-cycles, a contradiction since $S_4$ contains only six 2-cycles. Thus we see that $H = A_4$. $\square$

Lemma 2: If $H \leq S_4$ has order 6, then $H$ is maximal. Proof: If $H$ is not maximal, then $H$ is contained in some subgroup $K$ of order 12 by Lagrange. By Lemma 1, $K \cong A_4$, but $A_4$ contains no subgroups of order 6, a contradiction. Thus $H \leq S_4$ is maximal. $\square$

Now to the main results.

1. $\langle (1\ 2\ 3) \rangle$ and $\langle (1\ 2) \rangle$ are distinct maximal subgroups, since each has prime index. Thus $\Phi(S_3) \leq \langle (1\ 2\ 3) \rangle \cap \langle (1\ 2) \rangle = 1$, so that $\Phi(S_3) = 1$.
2. We know from the subgroup lattice of $A_4$ that $\langle (1\ 2\ 3) \rangle$ and $\langle (1\ 2\ 4) \rangle$ are distinct maximal subgroups of prime order. Thus $\Phi(A_4) = 1$.
3. Note that $H_1 = \langle (a\ b\ c), (a\ b) \rangle$ and $H_2 = \langle (a\ b\ d), (a\ d) \rangle$ are distinct subgroups of order 6 in $S_4$, which are maximal by Lemma 2. Evidently $H_1 \cap H_2 = \langle (a\ b) \rangle$. Now $A_4 \leq S_4$ is also maximal, and $(a\ b)$ is an odd permutation. Thus $H_1 \cap H_2 \cap A_4 = 1$, and we have $\Phi(S_4) = 1$.
4. $\Phi(A_5) \leq A_5$ is normal and proper. Since $A_5$ is simple, $\Phi(A_5) = 1$.
5. We found in a lemma to the previous exercise that $\Phi(S_5) = 1$.

Sym(4) is not a commutator subgroup

Prove that there is no group whose commutator subgroup is $S_4$.

Suppose $G$ is a group such that $G^\prime \cong S_4$. We computed in §5.4 #4 that $G^{\prime\prime} \cong S_4^\prime = A_4$ and $G^{\prime\prime\prime} \cong V_4$. Note that $G^\prime/G^{\prime\prime} \cong Z_2$ and $G^{\prime\prime}/G^{\prime\prime\prime} \cong Z_3$ are cyclic; by this previous exercise, we have $G^{\prime\prime} = G^{\prime\prime\prime}$, a contradiction.

Thus no such group $G$ exists.

Compute the upper and lower central series for Alt(n) and Sym(n) where n is at least 5

Find the upper and lower central series for $A_n$ and $S_n$ where $n \geq 5$.

1. First we consider $A_n$.

Recall that the higher centers of $G$ are characteristic, hence normal. Since $A_n$ is simple for $n \geq 5$, and $A_n$ is not abelian, we have $Z_0(A_n) = 1$ and $Z_1(A_n) = 1$. Thus the upper central series of $A_n$, $n \geq 5$, is $1$.

Now $A_n^0 = A_n$ and $A_n^1 = [A_n,A_n]$. Since $[(a\ b\ c), (a\ d\ e)] = (a\ c\ e)$, $A_n^1$ contains all 3-cycles, and thus $A_n \leq A_n^1$. Hence $A_n^1 = A_n$, and the lower central series of $A_n$ is $A_n$.

2. Now we consider $S_n$.

We have seen previously that $Z(S_n) = 1$ for $n \geq 5$. Thus Z_1(S_n) = Z_0(S_n)\$, and the upper central series of $S_n$ is $1$.

Now $S_n^0 = S_n$, and we saw in §5.4 #5 that $S_n^1 = S_n^\prime = A_n$. Now $A_n = [A_n,A_n] \leq [S_n,A_n] = S_n^2 \leq A_n$, so that the lower central series of $S_n$ is $S_n \geq A_n$.

Compute the upper and lower central series of Sym(4) and Alt(4)

Find the upper and lower central series of $S_4$ and $A_4$.

Recall that $G^\prime = [G,G] = G^1$.

1. First we consider $A_4$. Since $(a\ b)(c\ d)(a\ b\ c) = (d\ c\ b)$ and $(a\ b\ c)(a\ b)(c\ d) = (a\ c\ d)$, $Z(A_4) = 1$. Thus the upper central series of $A_4$ is $1$.

Now $A_4^0 = A_4$. In §5.4 #4, we found that $A_4^\prime = A_4^1 = V_4$. Using the computations above, we can see that $[A_4,V_4]$ contains all products of two 2-cycles and is contained in $V_4$; thus $[A_4,V_4] = V_4$. Thus the lower central series of $A_4$ is $A_4 \geq V_4$.

2. First we consider $S_4$. We have seen previously that $Z(S_4) = 1$; thus the upper central series of $S_4$ is $1$.

Now $S_4^0 = S_4$. We computed that $S_4^\prime = S_4^1 = A_4$. Now $[S_4,A_4]$ is contained in $A_4$; moreover, since $(a\ d)(a\ c\ b)(a\ d)(a\ b\ c) = (a\ d\ c)$, $[S_4,A_4]$ contains all 3-cycles, so that $A_4 \leq S_4^2$. Thus the lower central series of $S_4$ is $S_4 \geq A_4$.