## Tag Archives: alternating group

### Exhibit a presentation of Alt(4) using two generators

Establish a finite presentation for $A_4$ using two generators.

We know that $A_4$ is generated by an element of order 2 and an element of order 3. Let $\alpha = (1\ 2)(3\ 4)$ and $\beta = (1\ 2\ 3)$. Evidently, these generators satisfy the relations $\alpha^2 = \beta^3 = (\alpha\beta)^3 = 1$. To show that these relations provide a presentation for $A_4$, it suffices to show that the group $\langle \alpha, \beta \ |\ \alpha^2 = \beta^3 = (\alpha\beta)^3 \rangle$ has order at most 24. To that end, we will compute the reduced words in this group.

• There is one word of length 0: 1.
• There are two words of length 1: $a$ and $b$. Both are reduced by definition.
• There are at most four words of length 2: $a^2$, $ab$, $ba$, and $b^2$. Since $a^2 = 1$, it is not reduced. Thus 3 words of length 2 might be reduced.
• There are at most six words of length 3: $aba$, $ab^2$, $ba^2$, $bab$, $b^2a$, and $b^3$. Since $ba^2 = b$ and $b^3 = 1$, these words are not reduced. The remaining 4 words might be reduced.
• There are at most eight words of length 4: $aba^2$, $abab$, $ab^2a$, $ab^3$, $baba$, $bab^2$, $b^2a^2$, and $b^2ab$. Since $aba^2 = ab$, $abab = b^2a$, $ab^2a = bab$, $ab^3 = a$, $baba = ab^2$, and $b^2a^2 = b^2$, these words are not reduced. The remaining two words might be reduced.
• There are at most four words of length 5: $bab^2a = b^2ab$, $bab^3 = ba$, $b^2aba = bab^2$, and $b^2ab^2 = aba$. None of these are reduced.

Thus any group presented by $\langle a,b \ |\ a^2 = b^3 = (ab)^3 = 1 \rangle$ has cardinality at most 12, and we have the presentation $A_4 = \langle \alpha, \beta \ |\ \alpha^2 = \beta^3 = (\alpha\beta)^3 = 1 \rangle$.

### A group of order 12 with no subgroup of order 6 is isomorphic to Alt(4)

Show that a group of order 12 with no subgroup of order 6 is isomorphic to $A_4$.

Note that $12 = 2^2 \cdot 3$, 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. Let $x \in P_2$ have order 2 and $y \in P_3$ have order 3 by Cauchy; then $xy$ has order 6, a contradiction. Now suppose $n_2(G) = 3$. 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 $|N_G(P_2 \cap Q_2)|$ is divisible by $2^2$ and another prime; thus we have $P_2 \cap Q_2$ normal in $G$. Recall that every normal group of order 2 in a finite group is central; thus we have an element $x$ of order 2 in the center of $G$. Let $y \in G$ have order 3 by Cauchy; then $xy$ has order 6, a contradiction.

Thus $n_3(G) = 4$. Let $P_3 \leq G$ be a Sylow 3-subgroup and let $N = N_G(P_3)$. Now $N$ has index 4 and is self-normalizing by a previous result; in particular, $N$ is not normal in $G$. Now $G$ acts on $G/N$ by left multiplication, yielding a permutation representation $G \rightarrow S_{G/N}$ whose kernel $K$ is contained in $N$. Note that $|N| = 3$, so that either $K = 1$ or $K = N$. If $K = N$, then for all $g \in G$ and $h \in N$, we have $hgN = gN$, so that $g^{-1}hg \in N$. Thus, for all $g \in G$, $gNg^{-1} \leq N$; since $G$ is finite, this implies that $N \leq G$ is normal, a contradiction. Thus $K = 1$ and in fact $G \leq S_4$. Since $G$ contains $4 \cdot 2 = 8$ elements of order 3, it contains all elements of order 3 in $S_4$. The subgroup generated by the three cycles in $S_4$ is $A_4$, so that $A_4 \leq G$; since these subgroups are finite and have the same cardinality, in fact $G \cong A_4$.

### 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 $\Phi(S_5) = 1$.

### 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 $S_4 \geq A_4$.

### Exhibit a representative of each cycle type in Alt(5) as a commutator in Sym(5)

Exhibit a representative of each cycle type in $A_5$ as a commutator in $S_5$.

The cycle types in $A_5$ are $(5)$, $(3)$, and $(2,2)$.

1. $(a\ b)(c\ d)(a\ c\ e)(a\ b)(c\ d)(a\ e\ c) = (a\ b\ d\ e\ c)$
2. $(a\ c\ b)(a\ b)(a\ b\ c)(a\ b) = (a\ b\ c)$
3. $(a\ b)(c\ d)(a\ c\ b)(a\ b)(c\ d)(a\ b\ c) = (a\ d)(b\ c)$

### For n at least 5, the commutator subgroup of Sym(n) is Alt(n)

Prove that the commutator subgroup of $S_n$ is $A_n$ for all $n \geq 5$.

We know that $[S_n,S_n]$ is characteristic, hence normal, in $S_n$. Since $n \geq 5$, then, this commutator is either 1, $A_n$, or $S_n$.

Now $[S_n,S_n] \neq 1$ since $S_n$ is nonabelian. Moreover, since $S_n/A_n \cong Z_2$ is abelian, we have that $[S_n, S_n] \leq A_n$. Thus $[S_n,S_n] = A_n$.

### Compute the commutator subgroups of Sym(4) and Alt(4)

Find the commutator subgroups of $S_4$ and $A_4$.

We start with $S_4$. Note that $A_4$ is normal in $S_4$ and $S_4/A_4 \cong Z_2$ is abelian, so that by Proposition 7, $S_4^\prime \leq A_4$. Note that $(a\ c\ b)(a\ b)(a\ b\ c)(a\ b) = (a\ b\ c)$; that is, $(a\ b\ c) = [(a\ b\ c), (a\ b)]$. Now $S_4^\prime$ is a characteristic, hence normal, subgroup, and thus is a union of conjugacy classes. Since the conjugacy classes in $S_4$ are precisely the cycle shapes, all 3-cycles are in $S_4^\prime$.

Since $(a\ b)(c\ d)(a\ c\ b)(a\ b)(c\ d)(a\ b\ c) = (a\ d)(b\ c)$, all products of two 2-cycles are also in $S_4^\prime$. With the identity, this exhausts the elements of $A_4$; hence $S_4^\prime = A_4$.

Now we consider $A_4$. We know that $V_4 \leq A_4$ is characteristic, hence normal, and that (by Lagrange) $A_4/V_4 \cong Z_3$ is abelian. By Proposition 7, $A_4^\prime \leq V_4$. Note moreover that since $(a\ b\ c)(a\ b)(c\ d)(a\ c\ b) = (a\ d)(b\ c)$, every order 2 subgroup is not normal in $A_4$, hence not characteristic, and thus cannot be the commutator subgroup. Thus $A_4^\prime = V_4$.

### Construct an infinite simple group

Let $\Omega = \{1,2,3,\ldots\}$ (as in this previous exercise). Let $D$ be the subgroup of $S_\Omega$ consisting of permutations which move only a finite number of elements of $\Omega$ (described in this previous exercise) and let $A$ be the set of all elements $\sigma \in D$ that act as even permutations on the (finite) set $M(\sigma)$. Prove that $A$ is an infinite simple group. [Hint: show that every pair of elements of $A$ lie in a finite simple subgroup of $A$.]

We begin with a lemma.

Lemma: If $x,y \in A$, then there is a finite simple group $N \leq A$ with $x,y \in N$. Proof: If $x,y \in A$, then $M(x)$ and $M(y)$ are finite. Let $S = \{a_1,a_2,a_3,a_4,a_5\}$ be five elements in $\Omega$ not moved by either $x$ or $y$. Now $T = M(x) \cup M(y) \cup S$ is finite, and by Theorem 24 in the text, $A_T \leq A$ is simple. Clearly $x,y \in A_T$. $\square$

Now suppose $H \leq A$ is a normal subgroup. Suppose $H \neq 1$; let $x \in H$ be a nonidentity. Suppose that there exists $y \in A \setminus H$. Now there exists a finite simple group $K \leq A$ with $x,y \in K$. Moreover, $H \cap K \leq K$ is a nontrivial normal subgroup (containing in particular $x \neq 1$) so that, since $K$ is simple, $H \cap K = K$. Thus $K \leq H$. But then $y \in H$, a contradiction. Thus no such element $y$ exists, and we have $H = A$.

Thus if $H \leq A$ is a normal subgroup, either $H = 1$ or $H = A$. Hence $A$ is simple.

Finally, it is clear that $A$ is infinite since, for example, $A_n \leq A$ is a subgroup for every integer $n$.

### Alt(n) is generated by the set of all 3-cycles

Prove that $A_n$ is generated by the set of all 3-cycles for $n \geq 3$.

Let $n \geq 3$ and let $T = \{ (a\ b\ c) \ |\ 1 \leq a,b,c \leq n \}$ be the set of 3-cycles in $A_n$.

Note that $A_n$ contains $T$, so that $\langle T \rangle \leq A_n$.

Recall that $A_n$ consists of all permutations which can be written as an even product of transpositions; more specifically, $A_n$ is generated by the set of all products of two distinct 2-cycles. (Here we use the fact that $n \geq 3$.) Each product $\sigma$ of two 2-cycles has one of two forms.

If $\sigma = (a\ b)(c\ d)$, note that $\sigma = (a\ c\ b)(a\ c\ d)$.

If $\sigma = (a\ b)(a\ c)$, note that $\sigma = (a\ c\ b)$.

Thus $A_n \leq \langle T \rangle$, hence $A_n = \langle T \rangle$.