We’re all familiar with Cayley’s theorem: every group G is isomorphic to a subgroup of \text{Sym}(G). In the finite case, every finite group of order n is isomorphic to a subgroup of S_n. Now let’s see some applications of Cayley’s idea.

Problem 1. Let G be a group and let H be a subgroup of G with [G:H]=n < \infty. Prove that there exists a normal subgroup N of G such that N \subseteq H and [G:N] \mid n!.

Solution. Let X = \{x_1H, \cdots , x_n H \} be the set of all left cosets of H in G. Define

\varphi: G \longrightarrow \text{Sym}(X) \simeq S_n

by \varphi(g)(x_jH)=gx_jH, \ \forall g \in G, \forall \ 1 \leq j \leq n. Then see that \varphi(g) \in \text{Sym}(X). Thus \varphi is well-defined. It is easy to show that \varphi is a group homomorphism and N=\ker \varphi \subseteq H. So G/N is isomorphic to a subgroup of \text{Sym}(X). Thus [G:N] \mid |\text{Sym}(X)|=n!. \ \Box

Problem 2. Let G be a finite group and let p be the smallest prime divisor of |G|. Let H be a subgroup of G with [G:H]=p. Prove that H is a normal subgroup of G.

Solution. By Problem 1, there exists some normal subgroup N \subseteq H such that [G:N] \mid p!. Therefore p!|N|=kp|H|, for some integer k. Thus (p-1)!|N|=k|H|. But \gcd(|H|,(p-1)!)=1, because no prime divisor of |H| is smaller than p. So |H| \mid |N| and hence N=H because N \subseteq H. \ \Box

A trivial result of Problem 2 is that if H is a subgroup of a group G and [G:H]=2, then H is normal in G. Also if p is a prime number and G is a group of order p^n, then every subgroup of G of order p^{n-1} is normal.

Problem 3. Prove that if G is an infinite simple group and H is a proper subgroup of G, then [G:H]=\infty.

Solution. Suppose to the contrary that [G:H]=n < \infty. Then, by Problem 1, H  contains some normal subgroup N with [G:N] \mid n!. In particular [G:N] < \infty. But G is simple and N \neq G, because N \subseteq H. Hence N = \{1\} and so [G:N]=|G|=\infty, which is a contradiction! \Box

Problem 4. Let n \geq 5 and let H be a proper subgroup of A_n. Prove that [A_n:H] \geq n.

Solution. Let [A_n:H]=m. By Problem 1, A_n has a normal subgroup N which is contained in H. But A_n is simple because n \geq 5. Thus N = \{1\} and hence, by Problem 1, [A_n:N]=|A_n|=\frac{n!}{2} \mid m!, which is not possible unless m \geq n. \ \Box

Example. By Problem 4, A_6 has no subgroup of orders 72, \ 90, \ 120, \ 180.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s