Originally Posted by
Moo Hmm but yours is dealing with even/uneven, while we're talking about rational/irrational here ?
They may not be the same but they are similar.
In general:
Let $\displaystyle G$ be a group and $\displaystyle H$ be a subgroup. Then for any $\displaystyle x,y\in G,$ $\displaystyle x\in H$ and $\displaystyle y\notin H$ $\displaystyle \implies$ $\displaystyle xy\notin H.$
Proof is simple: Suppose $\displaystyle x$ (and therefore $\displaystyle x^{-1})\,\in\,H.$ If $\displaystyle xy\in H,$ then $\displaystyle y=x^{-1}(xy)\in H.$ Therefore by contrapositivity, $\displaystyle y\notin H\ \implies\ xy\notin H.$
In bmp05’s original example, $\displaystyle G$ is the addibive group of the reals and $\displaystyle H$ is the additive group of the rationals; hence rational + irrational = irrational. In the odd/even example, $\displaystyle G$ is the additive group of the integers and $\displaystyle H$ is the subgroup of the even integers – $\displaystyle \therefore$ even + odd = odd.
Further examples:
(i) Even permutation × odd permutation = odd permutation. (Take $\displaystyle G$ to be a symmetric group, $\displaystyle H$ to be the corresponding alternating group.)
(ii) Rotation × reflection = reflection. (Take $\displaystyle G$ to be a dihedral group, $\displaystyle H$ to be the subgroup of rotations.)