Problem 1A

Here are solutions to some of the exercises from Finite Group Theory by I. Martin Isaacs.

1A.1

Let \(H\) be a subgroup of prime index \(p\) in the finite group \(G\), and suppose that no prime smaller than \(p\) divides \(|G|\). Prove that \(H\triangleleft G\).

Proof.

Let \(\phi:G\rightarrow G/\mathrm{core}_G(H)\). From Theorem 1.1 it follows that \(\phi(G)\simeq A \leq S_n\), where \(n=|G:H|=p\). Since \(p\) is the minimal prime number dividing \(G\), then \(|A |=p^k\). From the fact that \(|S_p|\) is not divisible by \(p^2\) it follows that \(|A |=p\). Thus \(|G:H|=|G: \mathrm{core}_G(H)|\), and therefore \(|H|=|\mathrm{core}_G(H)|\). From the fact that \(\mathrm{core}_G(H)\leq H\) it follows that \(\mathrm{core}_G(H)=H\).

1A.2

Given subgroups \(H,K\subseteq G\) and an element \(g\in G\), the set \(HgK = \{hgk | h\in H, k\in K\}\) is called an \((H,K)\)-double coset. In the case where H and K are finite, show that \(|HgK|=|H||K|/|K\cap H^g|\).

Proof.

Let \(K\) act on \(\Omega = \{Hgk | k\in K\}\) by right multiplication. It is easy to see that the action is transitive on \(\Omega\).

Every element in \(\Omega\) is a right coset of \(H\), their orders are all equal to \(|H|\), and any two distinct elements are disjoint, and their union is \(HgK\), so \(|\Omega|=|HgK|/|H|\).

Now we consider the stabilizer of \(Hg\). For \(k\in K\), then \(Hgk=Hg\) if and only if \(k\in H^g\). Therefore the stabilizer of \(Hg\) is \(K\cap H^g\). Hence \(|HgK|/|H|=|K:K\cap H^g|\). It follows that \(|HgK|=|H||K|/|K\cap H^g|\).

1A.3

Suppose that \(G\) is finite and that \(H,K\in G\) are subgroups.

1. Show that \(|H : H \cap K| \leq |G : K|\), with equality if and only if \(HK = G\).

2. If \(|G : H|\) and \(|G : K|\) are coprime, show that \(HK = G\).

Proof.

1. Applying \(g=1\) in the previous problem, we obtain that \(|HK|=|H||K|/|H\cap K|\). Since \(HK\) is a subset of \(G\), \(|HK|\leq |G|\). Therefore \(|H : H \cap K| \leq |G : K|\). When equality holds, \(|HK|=|G|\), implying \(HK=G\).

2. Since \(G\) is a finite group, let \(|G|=p_1^{n_1}p_2^{n_2}\cdots p_r^{n_r}\), where \(p_1, p_2,\ldots, p_r\) are different primes. Since \(|G : H|\) and \(|G : K|\) are coprime, without loss of generality, we can let

\[|G:H|=p_1^{m_1}p_2^{m_2}\cdots p_s^{m_s}\]

and

\[|G:K|=p_t^{m_t}p_{t+1}^{m_{t+1}}\cdots p_r^{m_r},\]

where \(1\leq s<t\leq r\) and \(m_i\leq n_i\) for \(i=1,\ldots,s,t,\ldots,r\). Hence

\[|H|=p_1^{n_1 - m_1}\cdots p_s^{n_s - m_s} p_{s+1}^{n_{s+1}}\cdots p_{r}^{n_r}\]

and

\[|K|=p_1^{n_1}\cdots p_{t-1}^{n_{t-1}} p_{t}^{n_t - m_t}\cdots p_{r}^{n_r - m_r}.\]

By Lagrange theorem, \(|H\cap K|\) divides both \(|H|\) and \(|K|\). Hence \[|H\cap K\leq p_1^{n_1 - m_1}\cdots p_s^{n_s - m_s} p_{s+1}^{n_{s+1}}\cdots p_{t-1}^{n_{t-1}} p_{t}^{n_t - m_t}\cdots p_{r}^{n_r - m_r}.\] It is easy to see that \(|HK|=|H||K|/|H\cap K| \geq |G|\) by calculation. Therefore \(HK=G\).

Note. A more concise proof

1. \(|H|/|H\cap K|=|HK|/|K|\leq |G|/|K|=|G:K|\). Equality only in case \(HK=G\).

2. \(|G:H||G:K||H\cap K|\) divides \(|G|\). We have \(|G:H||G:K||H\cap K|=|G||G||H\cap K/(|H||K|)=|G|^2/|HK|\). Therefore \(|HK|=|G|\).

1A.4

Suppose that \(G=HK\), where \(H\) and \(K\) are subgroups. Show that also \(G=H^x K^y\) for all elements \(x,y\in G\). Deduce that if \(G=HH^x\) for a subgroup \(H\) and an element \(x\in G\), then \(H=G\).

Proof.

Since \(G^y=G\) and \((H^xK^y)^{y^{-1}}=H^{xy^{-1}}K\), it is enough to prove that \(G=H^xK\) for all \(x\in G\). Since \(x\in G=HK\), let \(x=hk\) where \(h\in H, k\in K\). Then \(H^xK=H^{hk}K=H^kK=k^{-1}HK=kG=G\).

1A.5

An action of a group \(G\) on a set \(\Omega\) is transitive if \(\Omega\) consists of a single orbit. Equivalently, \(G\) is transitive on \(\Omega\) if for every choice of points \(\alpha,\beta\in \Omega\), there exists an element \(g\in G\) such that \(\alpha \cdot g=\beta\). Now assume that a group \(G\) acts transitively on each of two sets \(\Omega\) and \(\Lambda\). Prove that the natural induced action of \(G\) on the cartesian product $$ is transitive if and only if \(G_{\alpha}G_{\beta} = G\) for some choice of \(\alpha \in \Omega\) and \(\beta \in \Lambda\).

Proof.

Let \((\alpha _1, \beta _1)\) \(\in\) \(\Omega \times \Lambda\). Since \(G\) acts transitively on each of two sets \(\Omega\) and \(\Lambda\), there exist two elements \(a,b\in G\) such that \(\alpha_1 =\alpha\cdot g\) and \(\beta_1 = \beta\cdot g\). If \(G_{\alpha}G_{\beta} = G\), then we can get that \(a=a_1 b_1\) and \(b=a_2 b_2\) where \(a_1, a_2\in G_{\alpha}, b_1,b_2\in G_{\beta}\). It is easy to confirm that \((\alpha_1, \beta_1)=(\alpha, \beta)\cdot a a_1^{-1}b b_2^{-1}\). Therefore \(G\) acts transitively on $$.

On the other hand, if \(G\) acts transitively on $$, we can show that \(G=G_{\alpha}G_{\beta}\) as follows. Let \(x\in G\). Assume that \((\alpha_2, \beta_2)=(\alpha,\beta)\cdot x\). If \(G\) acts transitively on $$, then there exists two elements \(u,v\in G\) such that \((\alpha,\beta_2)=(\alpha, \beta)\cdot u\), \((\alpha_2, \beta)=(\alpha, \beta)\cdot v\). It is clear that \(u\in G_{\alpha}\) and \(v\in G_{\beta}\). Moreover, \(uvx^{-1}\in G_{\alpha}\cap G_{\beta}\). Let \(uvx^{-1}=h\in G_{\alpha}\cap G_{\beta}\). Therefore \(x=(h^{-1}u)v\in G_{\alpha}G_{\beta}\). Thus \(G=G_{\alpha}G_{\beta}\).

1A.6

Let \(G\) act on \(\Omega\), where both \(G\) and \(\Omega\) are finite. For each element \(g\in G\), write \(\chi (g) = |\{\alpha \in \Omega | \alpha \cdot g = \alpha\}|\). The nonnegative-integer-valued function \(\chi\) is called the permutation character associated with the action. Show that $ {gG} (g)={} |G_{}|=n|G|,$ where \(n\) is the number of orbits of \(G\) on \(\Omega\).

Proof.

Consider \(\mathcal{A}=\{(\alpha, g)|\alpha \cdot g=\alpha\}\). There are two ways to count the number of elements in \(\mathcal{A}\). One is fixing \(\alpha\) and the other is fixing \(g\). Therefore we get that \(|\mathcal{A}|= \sum_{g\in G} \chi (g)=\sum_{\alpha \in \Omega} |G_{\alpha}|.\) If \(\mathcal{O}\subseteq \Omega\) is an orbit, then \(\sum_{\alpha \in \mathcal{O}} |G_{\alpha}|=|G|\). Therefore \(\sum_{\alpha \in \Omega} |G_{\alpha}|=n|G|\).

1A.7

Let \(G\) be a finite group, and suppose that \(H < G\) is a proper subgroup. Show that the number of elements of \(G\) that do not lie in any conjugate of \(H\) is at least \(|H|\).

Proof.

Let \(\chi\) be the permutation character associated with the right multiplication action of \(G\) on the right cosets of \(H\). Since the action is transitive, we can get \(\sum_{g\in G} \chi(g)=|G|\) from the previous problem. Consider the action of \(H\) on the right cosets of \(H\) by right multiplication. Since \(H<G\), this action is not transitive.

It follows that \(\sum_{h\in H} \chi (h)\geq 2|H|\). Let \(M=\bigcup _{x\in G} H^x\). Since \(Hxg=Hx\iff H^xg=H^x\iff g\in H^x\), \(\chi(g)=0\) if and only if \(g\notin M\). Therefore \(\sum_{g\in M} \chi(g)=|G|\). Hence

\[ |M| \leq \sum_{g\in M\setminus H} \chi (g) + |H| \leq \sum_{g\in M\setminus H} \chi (g) + 2|H| - |H| \\ \leq \sum_{g\in M\setminus H} \chi (g) + \sum_{h\in H} \chi (h) - |H| \leq \sum_{g\in M} \chi (g) - |H| = |G| - |H|. \]

This inequality completes the proof.

1A.8

Let \(G\) be a finite group, let \(n > 0\) be an integer, and let \(C\) be the additive group of the integers modulo \(n\). Let \(\Omega\) be the set of \(n\)-tuples \((x_1,x_2,\ldots,x_n)\) of elements of G such that \(x_1x_2\cdots x_n = 1\).

1. Show that \(C\) acts on \(\Omega\) according to the formula \[(x_1,x_2,\ldots,x_n)\cdot k=(x_{1+k},x_{2+k},\ldots,x_{n+k}),\] where \(k\in C\) and the subscripts are interpreted modulo \(n\).

2. Now suppose that \(n = p\) is a prime number that divides \(|G|\). Show that \(p\) divides the number of \(C\)-orbits of size \(1\) on \(\Omega\), and deduce that the number of elements of order \(p\) in \(G\) is congruent to \(-1\) mod \(p\).

Proof.

1. It clear that \(a,b\in G\) is permutable if \(ab=1\). Therefore

\[x_{1+k}x_{2+k}\ldots x_{n+k}=(x_m \ldots x_n)(x_1\ldots x_{m-1}) =(x_1\ldots x_{m-1})(x_m \ldots x_n)=1\]

where \(m\equiv 1+k \pmod{n}\) and \(1\leq m \leq n\). It's easy to verify that \[ (x_1,x_2,\ldots,x_n)\cdot 0= (x_1,x_2,\ldots,x_n) \] and \[ ((x_1,x_2,\ldots,x_n)\cdot k_1)\cdot k_2) =(x_1,x_2,\ldots,x_n)\cdot (k_1 k_2) .\] Hence \(C\) acts on \(\Omega\).

2. We can choose \(x_1, x_2, \ldots, x_{p-1}\) arbitrarily and \(x_p=(x_1 x_2\cdots x_{p-1})^{-1}\) is completely determined by the choices of these previous elements. Hence \(|\Omega|=|G|^{p-1}\). Let \(\mathcal{O}\) be a \(C\)-orbit. If \(|\mathcal{O}|=1\), then \(\mathcal{O}=\{(x,x,\ldots,x)\}\). It follows that \(x^p=1\). Then we consider the case when \(|\mathcal{O}|\neq 1\). Let \((x_1,x_2,\ldots, x_p)\in \mathcal{O}\) and \(x_1,x_2,\ldots,x_p\) are not all equal. Then the following \(p-1\) \(p\)-tuples \[(x_2,\ldots,x_p,x_1), (x_3,\ldots,x_p,x_1,x_2),\ldots,(x_p,x_1,\ldots x_{p-1})\] are also in \(\mathcal{O}\) and \[(x_1,\ldots, x_p)\neq (x_m,\ldots,x_p,x_1,\ldots,x_{m-1})\] if \(2\leq m\leq p\) (This can be derived from Fermat's little theorem). Hence in this case \(p\mid |\mathcal{O}|\). Since \(p\) divides \(|G|\), the sum of the orders of all orbits, \(\sum |\mathcal{O}|=|\Omega|=|G|^{p-1}\) is devided by \(p\). If we remove all orbits with more than one element, we get that the sum of the orders of the orbits with exactly one element, \(\sum_{|\mathcal{O}|=1} |\mathcal{O}|\), is also divisible by \(p\). Hence \(p\) divides the number of \(C\)-orbits of size \(1\) on \(\Omega\). Removing the identity element, the number of elements of order \(p\) is congruent to \(-1\) modulo \(p\).

1A.9

Suppose \(|G| = pm\), where \(p > m\) and \(p\) is prime. Show that \(G\) has a unique subgroup of order \(p\).

Proof.

Let \(M=\{g\mid g^p=1, g\neq 1\}\). By the previous problem, \(|M|\equiv -1 \pmod{p}\). Let \(|M|=kp-1\) where \(k\in \mathbb{N} ^*\). It's obvious that if \(h\in M\), then \(h^2,\ldots,h^{p-1}\in M\). Hence \(p-1 \mid |M|\). Since \(kp-1=k(p-1)+k-1\), we can get that \((p-1)\mid (k-1)\). Let \(k-1=l(p-1)\) where \(l\in \mathbb{N}\). Then \[|M|=(lp-l+1)(p-1)+l(p-1)=(lp+1)(p-1).\] Therefore \((lp+1)(p-1)\leq |G|=pm\). It follows that \(l=0\) by \(m<p\). Hence \(|M|=p-1\) and so \(G\) has a unique subgroup of order \(p\).

1A.10

Let \(H\subseteq G\).

1. Show that \(|\mathrm{N} _G(H):H|\) is equal to the number of right cosets of \(H\) in \(G\) that are invariant under right multiplication by \(H\).

2. Suppose that \(|H|\) is a power of the prime \(p\) and that \(|G:H|\) is divisible by \(p\). Show that \(|\mathrm{N} _G(H):H|\) is divisible by \(p\).

Proof.

1. Since \[HxH=Hx\iff H^x H=H^x\iff H^x=H \iff x\in \mathrm{N} _G(H),\] It's clear that (a) holds.

2. Let \(|H|=p^n\) and \(G=p^m\) where \(1\leq n<m\). Consider \(\Omega=\{HxH\mid x\in G\}\). It is clear that either \(HxH=HgH\) or \(HxH\cap HgH=\emptyset\). Hence \(|G|=\sum_{HxH\in \Omega} |HxH|\). By Problem 1A.2, \(|HxH|=|H|^2/|H\cap H^x|\). If \(|HxH|>|H|\), then \(p^{n+1}\mid |HxH|\). If \(|HxH|=|H|\), then \(H^x=H\) and so \(x\in \mathrm{N}_G(H)\). Since \(p^{n+1}\) divides \(|G|\), \(p^{n+1}\) divides \(|\mathrm{N}_G(H)|\). Hence by (a), \(|\mathrm{N} _G(H):H|\) is divisible by \(p\).