A fact on the interaction of subgroups with subgroups of prime index

Let G be a group, N \leq G a normal subgroup of prime index p, and K \leq G a subgroup. Prove that either K \leq N or G = NK and [K:K \cap N] = p.

Suppose K \setminus N \neq \emptyset; say k \in K \setminus N. Now G/N \cong \mathbb{Z}/(p) is cyclic, and moreover is generated by any nonidentity- in particular by \overline{k}.

Now KN \leq G since N is normal. Let g \in G. We have gN = k^aN for some integer a. In particular, g = k^an for some n \in N, hence g \in KN. We have [K:K \cap N] = p by the Second Isomorphism Theorem.

Post a comment or leave a trackback: Trackback URL.


  • Jer  On November 14, 2010 at 8:57 pm

    Can you please explain this? I really don’t understand how this proves what it’s asking to prove. Thank You.

    • nbloomf  On November 15, 2010 at 9:23 am

      Supposing K \not\leq N, we find an element k \in K such that kN generates G/N \cong Z_p. Then we find that every element of G has the form k^an for some integer a and element n \in N; hence G = KN. The index condition follows from the Second Isomorphism Theorem, since p = [KN:N] = [K : K \cap N].

      • Jer  On November 15, 2010 at 1:29 pm

        Why do you show G=KN and not G=NK? And also, why does p=|KN:N|? Thanks

        • Jer  On November 15, 2010 at 1:32 pm

          Sorry, I got the index is because |KN:N|=|G/N|=p.

  • Jer  On November 15, 2010 at 1:34 pm

    Sorry, I’m an idiot, KN=NK since N is normal, right?

    • nbloomf  On November 15, 2010 at 3:15 pm


  • Gobi Ree  On November 28, 2011 at 2:37 am

    Another approach to show G=NK: Suppose K \not\leq N, then \langle N,K \rangle = NK \supsetneq N so that NK/N is a nontrivial subgroup of G/N. So NK/N=G/N and by the lattice isomorphism theorem, G=NK.

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

%d bloggers like this: