My question is motivated by this paper in whose second page the author describes an analog of even and odd numbers for any additive group $G$ with a subgroup of index $2$, denoted $G^+$: the elements of this subgroup are called even elements and the elements of the complement, $G^-$, are called odd elements.
By additive group it is meant abelian, but I'm not sure if it must be cyclic.
The question is
Is the sum of two odd elements always an even element?
I tried to prove it doing the same that I would do for the integers. Namely, given two odd integers I can write them as $2n+1$ and $2m+1$ respectively, so their sum equals $2(m+n+1)$.
The analogy I tried was considering the cosets $\{G^+, G^-\}$ and letting $h_1,h_2\in G^-$ such that $h_1+h_2\notin G^+$. I can write them as $h_1=h_1'+g_1$, $h_2=h_2'+g_2$, where $h'_i\in G^-$ and $g_i\in G^+$. Here, $g_1$ and $g_2$ play the role of $2n$ and $2m$, respectively. The problem is that I can't find an analog to the number $1$.
If I assume $G$ to be cyclic, say $G=\langle g\rangle $, it would suffice to show that $g+g\in G^+$. However, I haven't been able to prove it. Are there example of (cyclic or non-cyclic) groups with a subgroup of order $2$ in which the statement is false or can I prove it for some kind of group?
Best Answer
In fact you can even drop the assumption that the group is abelian.
The starting point is that a subgroup $G^+<G$ of index $2$ is always normal (this comes at once if you try to write down explicitely the partitions into right and left cosets).
Thus the quotient group $G/G^+$ exists and it has $2$ elements, and so is isomorphic to $\Bbb Z/2\Bbb Z$.
From this it follows immediately that a product $g_1g_2$ is in $G^+$ if and only if both or none of the $g_i$ are in $G^+$.