Normal subgroup implies $g^2 \in H$ for any element $g$

group-theorynormal-subgroups

This question is exercise 2.16 of An Introduction to the Theory of Groups by Rotman.

2.16 If $H \le G$ has index $2$, then $g^2 \in H$ for every $g \in G$.

I know that the index being $2$ implies that $H$ is a normal subgroup. However, I don't quite see how that helps solve the problem. Also, this exercise comes before normal subgroups are even introduced, so it might be a motivating question for normal subgroups, or might not even require that fact.

I do know that by Lagrange's theorem:

$$|G|=|H| \cdot |G:H|$$
$$|G|= 2 |H|$$

and so the subgroup is necessarily exactly half the size of $G$. Also, this clearly implies that $G$ has an even number of elements and so I can make a statement regarding the parity of odd order elements and even order elements, however, I don't think that such a statement would be useful.

So, how should the statement be proved? The only guess that I have at this point is that it has to do with the fact that cosets partition the group and that this, combined with the above result via Lagrange's theorem, could possibly give the statement. Thoughts?

Best Answer

If you want to prove this without using quotients and normal subgroups, let $x,y\in G$.

  1. If $x,y\in H$, then $xy\in H$.
  2. If $x\in H$, $y\notin H$, then $xy\notin H$.
  3. If $x\notin H$, $y\in H$, then $xy\notin H$.
  4. If $x\notin H$ and $y\notin H$, where is $xy$? If $xy\notin H$, then $xy\in xH$, so there exists $h\in H$ such that $xy=xh$; but that means that $y=h\in H$, a contradiction. So $xy\in H$.

In particular: if $x\in G$, either $x\in H$ so $x^2\in H$, or $x\notin H$ so $xx=x^2\in H$.

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