2

True/False:
Every finite abelian group of even order has a subgroup of index 2.

There exists an element of order 2, and hence a subgroup order 2 (Call it $H$). Let H = Ker$\phi$
where $\phi:$ $G \rightarrow G/H$.
I was trying to use the fact that $G/H$ is isomorphic to a group of size $|G|/2$ by the Fundamental Homomorphism Theorem, combined with the fact that every subgroup of $G$ is normal. Does anyone know what piece(s) I'm missing?

sobrio35
  • 111
  • But $G/H$ is not a subgroup of $G$, it's a quotient of $G$. (Although every quotient happens to be isomorphic to a subgroup when $G$ is abelian.) – anon Mar 27 '16 at 03:21
  • @user326188 So use fact $\exists$ direct product size [G:H], from FTFGAG? – sobrio35 Mar 27 '16 at 03:23
  • @user43687 I'm looking for index, though, not order – sobrio35 Mar 27 '16 at 03:26
  • @sobrio35 I think I misunderstood your approach. I thought you were trying to identify the order 2 subgroup as having index two. You were trying to realize the quotient as a subgroup right? In any case, I think your going to need the structure theorem. –  Mar 27 '16 at 03:45

1 Answers1

1

You're not using the fact that $G$ is abelian (this is essential, since the claim is false for $G = A_5$, for example). Use the structure theorem for abelian groups.