Talk:Abelian Group of Order Twice Odd has Exactly One Order 2 Element

I see your result but it does not prove there is only one such element.

My idea (which came to me in the car the other day) was to suggest that if there are 2 such elements then there must be 2 such subgroups and their inner product would be a subgroup order 4, which by Lagrange contradicts the order twice odd condition. --prime mover (talk) 05:39, 28 September 2012 (UTC)


 * That was what I envisaged first as well, but if $g \ne h \ne e$ are two proper such elements, one would have that $\{g, h, e\} \subseteq H_2$, which is an apparent contradiction with the statement on Abelian Group Factored by Prime that $|H_2| = 2$. I employed the fact that $p^n = 2^1$ in this case. So I contend that it does prove that there is precisely one proper element with this property. --Lord_Farin (talk) 08:19, 28 September 2012 (UTC)
 * Of course, it may be worthwhile to produce the proof based on Lagrange as well; it appears on second reading that it gives a genuinely different approach. --Lord_Farin (talk) 08:22, 28 September 2012 (UTC)