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Homework Statement
Let G be a group of order 2n. Show that G contains an element of order 2. If n is odd and G is abelian, then there is only one element of order 2
Homework Equations
Theorem (Lagrange):
If H is a subgroup of a group G, then |G|=[G<img src="/styles/physicsforums/xenforo/smilies/arghh.png" class="smilie" loading="lazy" alt=":H" title="Gah! :H" data-shortname=":H" />]|H|. In particular, if G is finite, then the order of a divides G for all a \in G
The Attempt at a Solution
Case 1. G is cyclic.
This means that G \cong Z_{2n}.
We know that n + n = 0 for n \in Z_{2n}.
Thus there is an element of order 2 in G.
Case 2. G is not cyclic.
This is where I'm stuck.
I also don't know how to show the next statement when G is abelian and n is odd.