Theorem 8. Lagrange’s Theorem

If \(G\) is a finite group and \(H\) is a subgroup of \(G\), then the order of \(H\) divides the order of \(G\), and the number of left cosets of \(H\) in \(G\) equals \(\frac{|G|}{|H|}\).

Proof

Let \(k\) be the number of left cosets of \(H\) in \(G\).

For an arbitrary \(g \in G\), define \(\varphi:H \rightarrow gH\) by \(g \mapsto gh\). By definition of coset, \(\varphi\) is surjective. Also, \(gh_1 = gh_2 \implies h_1 = h_2\), and thus \(\varphi\) is injective. Therefore, $$|gH| = |H|$$

By Proposition 4, the set of left cosets of \(H\) in \(G\) partitions \(G\). Therefore, \(|G| = k|H|\) or \(k = \frac{|G|}{|H|}\).

Hence, \(|H|\) divides \(|G|\) and the number of left cosets of \(H\) in \(G\) equals \(\frac{|G|}{|H|}\).

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

Index

If \(G\) is a group (possibly infinite) and \(H \le G\), the number of left cosets of \(H\) in \(G\) is called the index of \(H\) in \(G\) and is denoted by \(|G:H|\).

Corollary 9

If \(G\) is a finite group and \(x \in G\), then the order of \(x\) divides the order of \(G\). In particular, \(x^{|G|} = 1\) for all \(x \in G\).

Proof

Let \(H = \langle x \rangle\). By Proposition 2.2, \(|H| = |x|\). Since \(H\) is a subgroup of \(G\), by Lagrange’s Theorem, \(|G| = n|H| = n|x|\) for some integer \(n\). Therefore, \(|x| \mid |G|\). In particular, $$x^{|G|} = x^{n|x|} = (x^{|x|})^n = 1^n = 1.$$

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

Corollary 10

If \(G\) is a finite group of prime order \(p\), then \(G\) is cyclic, hence \(G \cong Z_p\).

Proof

Since \(|G| = p > 1\), there exists \(x \in G\) such that \(x \ne 1\). Let \(H = \langle x \rangle\). Since \(H\) is a subgroup of \(G\), by Lagrange’s Theorem, \(|H| \mid |G|\). Since \(|G|\) is prime, \(|H|\) must equal \(|G|\). Therefore, \(G = H = \langle x \rangle\). By Theorem 2.4, \(G \cong Z_p\).

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

Theorem 11. Cauchy’s Theorem

If \(G\) is a finite group and \(p\) is a prime dividing \(|G|\), then \(G\) has an element of order \(p\).

Proof

See 4.5.03.

Theorem 12. Sylow’s Theorem

If \(G\) is a finite group of order \(p^{\alpha}m\), where \(p\) is a prime and \(p\) does not divide \(m\), then \(G\) has a subgroup of order \(p^{\alpha}\).

Proof

See Theorem 4.18.

Product of subgroups

Let \(H\) and \(K\) be subgroups of a group. Define $$HK = \{hk \mid h \in H, k \in K\}.$$

Proposition 13

If \(H\) and \(K\) are finite subgroups of a group, then $$ |HK| = \frac{|H||K|}{|H \cap K|}. $$

Proof

Notice \(HK\) is a union of left cosets. $$HK = \bigcup\limits_{h\in H}hK.$$

By Proposition 4, any pair of left cosets of \(K\) are either distinct or identical. As shown in the proof of Lagrange’s Theorem, each left coset of \(K\) is of size \(|K|\). Therefore, \(|HK| = n|K|\) where \(n\) is the number of distinct left cosets of \(K\).

By Proposition 4, for \(h_1, h_2 \in H\), \(h_1K = h_2K \iff h_2^{-1}h_1 \in K\). Therefore, $$h_1K = h_2K \iff h_2^{-1}h_1 \in H \cap K \iff h_1(H \cap K) = h_2(H \cap K)$$ It follows that the number of distinct left cosets of \(K\) equals that of \(H \cap K\).

Note that since \(H\) and \(K\) are groups, \(H \cap K\) is a subgroup of \(H\). By Lagrange’s Theorem, the number of left cosets of \(H \cap K\) is \(\frac{|H|}{|H \cap K|}\). Therefore, \(n = \frac{|H|}{|H \cap K|}\).

Hence, \(|HK| = n|K| = \frac{|H||K|}{|H \cap K|}\).

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

Proposition 14

If \(H\) and \(K\) are subgroups of a group, \(HK\) is a subgroup if and only if \(HK = KH\).

Proof

First, assume \(HK\) is a subgroup. Notice \(K = 1\circ K \le HK\) and \(H = H\circ 1 \le HK\). Therefore, because of the closure property of \(HK\), for any \(h \in H\) and \(k \in K\), \(kh \in HK\). That is, \(KH \subseteq HK\).

Additionally, note the following 3 facts.

• Since \(HK\) is a subgroup, there exist \(h_1 \in H\) and \(k_1 \in K\) such that \(hk = (h_1k_1)^{-1}\).
• Since \(K\) is a subgroup, there exists \(k_2 \in K\) such that \(k_1^{-1} = k_2\).
• Since \(H\) is a subgroup, there exists \(h_2 \in H\) such that \(h_1^{-1} = h_2\).

Therefore, $$ hk = (h_1k_1)^{-1} = k_1^{-1}h_1^{-1} = k_2h_2 \in KH $$ and thus \(HK \subseteq KH\).

Conversely, assume \(HK = KH\). Let \(a,b\) be arbitrary elements of \(HK\). We can write \(a = h_1k_1\) and \(b = h_2k_2\) for some \(h_1, h_2 \in H\) and \(k_1, k_2 \in K\). Note the following 3 facts.

• Since \(K\) is a group, there exists \(k_3 \in K\) such that \(k_3 = k_1k_2^{-1}\).
• Since \(H\) is a group, there exists \(h_3 \in H\) such that \(h_3 = h_2^{-1}\).
• Since \(HK = KH\), there exist \(h_4 \in H\) and \(k_4 \in K\) such that \(k_3h_3 = h_4k_4\).

Therefore,

$$ ab^{-1} = h_1k_1(h_2k_2)^{-1} = h_1k_1k_2^{-1}h_2^{-1} = h_1k_3h_3 = h_1h_4k_4 \in HK $$

Hence, by the Subgroup Criterion, \(HK\) is a subgroup.

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

Corollary 15

If \(H\) and \(K\) are subgroups of \(G\) and \(H \le N_G(K)\), then \(HK\) is a subgroup of \(G\). In particular, if \(K \trianglelefteq G\), then \(HK \le G\) for any \(H \le G\).

Proof

Since \(H \le N_G(K)\), \(hKh^{-1} = K\) for all \(h\in H\).

That is, for any \(h \in H\) and \(k_1 \in K\), there exists \(k_2 \in K\) such that \(hk_1h^{-1} = k_2\), or \(hk_1 = k_2h \in KH\). Thus, \(HK \subseteq KH\).

Similarly, for any \(h \in H\) and \(k_2 \in K\), there exists \(k_1 \in K\) such that \(hk_1h^{-1} = k_2\), or \(k_2h = hk_1 \in HK\). Thus, \(KH \subseteq HK\).

By Proposition 14, \(HK = KH \implies HK \le G\). Hence, \(HK\) is a subgroup.

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

Normalize, Centralize

• If \(A\) is any subset of \(N_G(K)\), we say \(A\) normalizes \(K\).
• If \(A\) is any subset of \(C_G(K)\), we say \(A\) centralizes \(K\).