# Why must $a$ be in one of these two cosets: $H, Ha$? [duplicate]

If $$H \le G$$ and $$H$$ has index 2 in $$G$$ then $$a^2 \in H, \forall a\in G$$
My try:

• If $$a\in H$$ then $$a^2 \in H$$ from closure.
• If $$a \notin H$$ then we have $$2$$ Left\Right Cosets: $${H, Ha}$$. Suppose that $$a^2 \in Ha$$ so that there is $$h \in H$$ so that $$a^2 = ha$$ and then $$a=h\in H$$ and it's contradiction so $$a^2$$ must be in $$H$$.

My question is, why must $$a$$ be in one of these two cosets: $$H, Ha$$?

• The disjoint union of all distinct cosets of a subgroup $H$ is the entire group $G$ so each element must fall in one of the distinct cosets. Nov 30, 2021 at 20:09
• The question has changed. Nov 30, 2021 at 20:18
• In answer to the question at the end, $a$ necessarily belongs to the coset $Ha$, since $e\in H$. (Or did you mean to ask why $a^2$ must belong to one of the two cosets?) Nov 30, 2021 at 20:20
• @BarryCipra But what about $a^2$' why it belongs to $H$ or $Ha$?
– Xavi
Nov 30, 2021 at 20:22
• @Xavi, see podiki's comment. My comment merely addressed the question as asked. If you edit the question (and ping me), I'll delete the comment. Nov 30, 2021 at 20:24

HINT: Note that $$H$$ and $$Ha$$ cannot intersect, otherwise there would be a $$h,h' \in H$$ such that $$h' = ha$$, which would imply $$h^{-1}h'=a$$, which would imply $$a \in H$$.
• And so to finish then, note that $a^2$ must be in $H$ if $a$ is not in $H$. Indeed from the above, $H$ and $Ha$ partition $G$. So if $a^2$ is not in $H$, then $a^2$ would have to be in $Ha$, which would imply that there is an $h \in H$ such that $a^2 = ha$, which, right-mulitplying both sides by $a^{-1}$, would give $a = h$. This contradicts $a \not \in H$.
Remember that the cosets are exactly the equivalence classes of the equivalence relation $$a\sim b\iff b^{-1}a\in H$$. We know that if we define an equivalence relation on a set $$X$$, the disjoint union of the equivalence classes is exactly $$X$$. So $$G=H\cup aH$$. Since $$a^2\in G$$, it must be in one of them.
$$H$$ has index $$2=600/300$$ in $$G$$, hence it is normal in $$G$$ and $$G/H$$ is isomorphic to the group of order $$2$$, hence for every $$a\in G$$, $$a^2H= H$$, so $$a^2\in H$$.