A group is not the union of conjugates of any proper subgroup

Assume H is a proper subgroup of the finite group G. Prove that G \neq \bigcup_{g \in G} gHg^{-1}; i.e., G is not the union of the conjugates of any proper subgroup.


There exists a maximal subgroup M containing H. If M is normal in G, then \bigcup_{g \in G} gHg^{-1} \subseteq \bigcup_{g \in G} gMg^{-1} = M \neq G. If M is not normal, we still have \bigcup_{g \in G} gHg^{-1} \subseteq \bigcup_{g \in G} gMg^{-1}. By the previous exercise, \bigcup_{g \in G} gMg^{-1} contains at most (|M|-1)[G : M] nonidentity elements. Thus |\bigcup_{g \in G} gHg^{-1}| \leq |G| - [G : M] + 1 < |G|, since [G:M] \geq 2.

In particular, because G is finite, G \neq \bigcup_{g \in G} gHg^{-1}. Thus G is not the union of all conjugates of any proper subgroup.

Post a comment or leave a trackback: Trackback URL.

Comments

  • C  On August 28, 2011 at 12:48 pm

    in the second to last sentence, (the middle term of the inequality), do you mean
    lGl – [G:M] +1 instead of [g:M]?

    • nbloomf  On September 6, 2011 at 10:04 am

      Yes. 🙂 Thanks!

  • Gobi Ree  On December 12, 2011 at 12:26 am

    The phrase “because G is finite” seems unnecessary. The finiteness was used for the formula in the previous exercise.

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: