# Tough Problem from Rotman's Group Theory, 3rd Ed

• Sep 30th 2009, 11:15 AM
signaldoc
Tough Problem from Rotman's Group Theory, 3rd Ed
Given a group G of order (2^m)k, k odd. If G contains an element of order 2^m, then (show that) the set of elements of G having odd order is a (normal) subgroup of G.

This is from Rotman, 3rd ed, p.46. He gives the following HINT: Consider G as permutations via Cayley's Th. and show that it contains an odd permutation.

I have gotten to the following facts:
The postulated element of order 2^m is an odd permutation, since it is regular and each cycle has order 2^m (so each cycle is odd) and no letter is fixed (so there are k cycles, k odd). Thus the group contains both odd permutions and even ones. Since these are exchanged by applying the odd permutation, the even and odd are equinumerous. The odd permuations are of even order, so the odd ordered elements, if any, are a proper subset. This subset is invariant, even characteristic, but it is not clear that it is closed under group operation (ie, a subgroup) since the group may not be abelian...

Thanks,
Gary

PS: This comes early in the book, before Sylow and other powerful results. Therefore, I assume there is an "elementary" argument....
• Sep 30th 2009, 07:27 PM
NonCommAlg
Quote:

Originally Posted by signaldoc
Given a group G of order (2^m)k, k odd. If G contains an element of order 2^m, then (show that) the set of elements of G having odd order is a (normal) subgroup of G.

This is from Rotman, 3rd ed, p.46. He gives the following HINT: Consider G as permutations via Cayley's Th. and show that it contains an odd permutation.

I have gotten to the following facts:
The postulated element of order 2^m is an odd permutation, since it is regular and each cycle has order 2^m (so each cycle is odd) and no letter is fixed (so there are k cycles, k odd). Thus the group contains both odd permutions and even ones. Since these are exchanged by applying the odd permutation, the even and odd are equinumerous. The odd permuations are of even order, so the odd ordered elements, if any, are a proper subset. This subset is invariant, even characteristic, but it is not clear that it is closed under group operation (ie, a subgroup) since the group may not be abelian...

Thanks,
Gary

PS: This comes early in the book, before Sylow and other powerful results. Therefore, I assume there is an "elementary" argument....

prove it by induction over $m$: if m = 0, there's nothing to prove. now let $G$ be a group of order $n=2^mk,$ with $k$ odd and $m \geq 1.$ let $g \in G$ be such that $o(g)=2^m$ and $H$ be the set of elements

of odd order in $G.$ let $G_1=G \cap A_n$ and $g_1=g^2.$ we have $o(g_1)=2^{m-1}.$ looking at $G$ as a subgroup of $S_n$ (Cayley's theorem), we'll see that the length of each cycle in the decomposition of an

element $x \in G$ is equal to $o(x).$ so $H \subseteq G_1$ and $g_1 \in G_1.$ also $[G:G_1]=2$ because $\forall x,y \in G-A_n: \ xy^{-1} \in G_1.$ hence $|G_1|=2^{m-1}k$ and the proof is complete by induction.
• Oct 1st 2009, 07:12 AM
signaldoc
Many thanks to NonCommAlg
Nicely done... My problem may have been in (obstinately) not believing the result in the first place.