Proving Properties of a Group with Every Element of Order 2

fk378
Messages
366
Reaction score
0

Homework Statement


Let G be a group such that every x in G\{e} has order 2.
(a) Let H<=G be a subgroup. Show that for every x in G the subset H U (xH) is also a subgroup.
(b) Show that if G is finite, then |G|=2^n for some integer n.


The Attempt at a Solution


For (a), I know that since x is also part of G, then if multiplied to H (left coset) it will also still be contained in G, but I don't know how to prove it.
For (b), since the order of every element in G is 2, then every set in G has 2 elements, so the order of G is just 2(elements) times how many sets there are...also having trouble proving this.

Please let me know if my train of thoughts are right.
 
Physics news on Phys.org
Hint for (a): Notice that every element is its own inverse.

Hint for (b): Use part (a).
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top