Even and odd elements in more general additive groups

Question

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?

Answer 1

Since $G^+$ has index $2$, $G/G^+$ is a group of order $2$, so it is isomorphic to $\mathbb{Z}/2\mathbb{Z}$. In particular, the non-identity element of $G/G^+$ is $G^-$, and thus in the quotient group $G^-+G^-=G^+$.

Now given any two odd elements $h_1,h_2$, we have $$(h_1+h_2)+G^+=(h_1+G^+)+(h_2+G^+)=G^-+G^-=G^+,$$ hence $h_1+h_2\in G^+$.

Note that since a group of index $2$ is necessarily normal, this works even in a non-abelian group; in fact this is exactly what happens with the even and odd permutations in the symmetric groups.

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This can also be proved without using quotient groups : if $h_1,h_2\in G^-$, then we also have $-h_1\in G^-$, as otherwise we would have $-h_1\in G^+$ and then $h_1\in G_+$. In particular, $h_2\in G^-=(-h_1)+G^+$, thus $h_2=-h_1+g$ for some $g\in G^+$; and then $h_1+h_2=g\in G^+$. Again, this works even if $G$ is not abelian.

Answer 2

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 $G^+$.