Theorem 3

Let \(G\) be a group, let \(H\) be a subgroup of \(G\), and let \(G\) act by left multiplication on the set \(A\) of left cosets of \(H\) in \(G\). Let \(\pi_H\) be the associated permutation representation afforded by this action. Then,

1. \(G\) acts transitively on \(A\)
2. the stabilizer in \(G\) of the point \(1H \in A\) is the subgroup \(H\)
3. the kernel of the action (i.e., the kernel of \(\pi_H\)) is \(\cap_{x \in G}\ xHx^{-1}\), and \(\ker \pi_H\) is the largest normal subgroup of \(G\) contained in \(H\).

Proof

(1)

We can let \(aH\) and \(bH\) be arbitrary two elements of \(A\), where \(a, b \in G\). We have

$$ aH = (ab^{-1}b)H = (ab^{-1}) \cdot bH. $$

Therefore, \(G\) acts transitively on \(A\).

$$\tag*{$\blacksquare$}$$

(2)

By definition, the stabilizer in \(G\) of \(1H \in A\) is $$ G_{1H} = \{g \in G \mid g \cdot 1H = 1H\} = \{g \in G \mid gH = H\} = H. $$

$$\tag*{$\blacksquare$}$$

(3)

By definition,

$$ \ker\pi_H = \{g \in G \mid gxH = xH\;\forall\ x \in G\} $$

Note $$ gxH = xH\;\forall x \in G \iff x^{-1}gxH = H\;\forall x \in G \iff x^{-1}gx \in H\;\forall x \in G. $$

Also note that, for arbitrary \(x \in G\) and for some \(h \in H\),

$$ x^{-1}gx \in H \iff x^{-1}gx = h \iff g = xhx^{-1} \iff g \in xHx^{-1}. $$

Therefore, \(\ker\pi_H = \{g \in G \mid g \in xHx^{-1}\;\forall x \in G\} = \bigcap\limits_{x \in G} xHx^{-1}\).

$$\tag*{$\blacksquare$}$$

First, we show that \(\ker\pi_H\) is a subgroup of \(G\). Note that \(\ker\pi_H = \bigcap\limits_{x \in G} xHx^{-1} \subseteq 1H1^{-1} = H\). Any \(a,b \in \ker\pi_H\) can be written as \(a = gh_1g^{-1}\) and \(b = gh_2g^{-1}\) for some \(h_1, h_2 \in H\) and \(g \in G\). Then it follows that $$ ab^{-1} = (gh_1g^{-1})(gh_2^{-1}g^{-1}) = gh_1h_2^{-1}g^{-1} \in gHg^{-1} \subseteq \ker\pi_H. $$ Therefore, by the Subgroup Criterion, \(\ker\pi_H\) is a subgroup of \(H\), and thus it’s also a subgroup of \(G\).

Next, we show that \(\ker\pi_H\) is a normal subgroup of \(G\). For any \(g \in G\), any \(k \in \ker\pi_H\) can be written as \(k = ghg^{-1}\) for some \(h \in H\). It follows that, for any \(g \in G\) and any \(k \in \ker\pi_H\), $$ gkg^{-1} = g(ghg^{-1})g^{-1} = (gg)h(g^{-1}g^{-1}) \in (gg)H(g^{-1}g^{-1}) \subseteq \ker\pi_H. $$ Therefore, \(g(\ker\pi_H)g^{-1} \subseteq \ker\pi_H\), and thus by Theorem 3.6, \(\ker\pi_H \trianglelefteq G\).

Finally, suppose \(N\) is a normal subgroup of \(G\) contained in \(H\). Then, \(N = gNg^{-1} \le gHg^{-1}\) for all \(g \in G\). Therefore,

$$ N \le \bigcap\limits_{x \in G}xHx^{-1} = \ker\pi_H. $$

$$\tag*{$\blacksquare$}$$

Corollary 4. Cayley’s Theorem

Every group is isomorphic to a subgroup of some symmetric group. If \(G\) is a group of order \(n\), then \(G\) is isomorphic to a subgroup of \(S_n\).

Proof

Let \(H = 1\) and apply Theorem 3 to obtain \(\pi_H\). Note we are considering an action by left multiplication on the set of left cosets of the identity subgroup \(H\). \(\pi_H\) is a homomorphism from \(G\) to \(S_G\).

By Theorem 3, \(\ker\pi_H = \bigcap\limits_{x \in G}xHx^{-1} = \bigcap\limits_{x \in G}x1x^{-1} = 1\). This means \(\pi_H\) is injective because for \(a,b \in G\),

$$ \pi_H(a) = \pi_H(b) \implies 1 = \pi_H(a)\pi_H(b)^{-1} = \pi_H(ab^{-1}) \implies ab^{-1} = 1 \implies a = b. $$

Therefore, \(\pi_H\) is a bijection between \(G\) and its image in \(S_G\). That is, \(G\) is isomorphic to a subgroup of \(S_G\).

$$\tag*{$\blacksquare$}$$

Corollary 5

If \(G\) is a finite group of order \(n\) and \(p\) is the smallest prime dividing \(|G|\), then any subgroup of index \(p\) is normal.

Proof

Suppose \(H \le G\) and \(|G : H| = p\). Let \(\pi_H\) be the permutation representation afforded by multiplication on the set of left cosets of \(H\) in \(G\), let \(K = \ker \pi_H\), and let \(|H : K| = k\). Then \(|G : K| = |G : H||H : K| = pk\).

By the First Isomorphism Theorem, \(G/K = G/ \ker \pi_H \cong \pi_H(G)\). Since \(H\) has \(p\) left cosets, \(\pi_H(G) \le S_p\). Since \(|S_p| = p!\) and \(\pi_H(G) \cong G/K\), by Lagrange’s Theorem, \(pk = |G/K|\) must divide \(p!\). It follows that \(k \mid (p-1)!\).

Note \(k \mid |G|\) because, by Lagrange’s Theorem, \(\frac{|G|}{|K|} = |G : K| = pk\). Additionally, \(p\) is the smallest prime dividing \(|G|\). Therefore, any primes that compose \(k\) are greater than or equal to \(p\). However, all prime divisors of \((p-1)!\) are smaller than \(p\), so \(k \mid (p-1)!\) forces \(k\) to be \(1\).

Therefore, \(|H:K| = k = 1\) and thus \(H = K\). Then, by Theorem 3 above, \(K\) is a normal subgroup of \(G\). Hence, \(H = K \trianglelefteq G\). $$\tag*{$\blacksquare$}$$