0
$\begingroup$

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: $H, Ha$?

$\endgroup$
6
  • $\begingroup$ 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. $\endgroup$
    – podiki
    Nov 30, 2021 at 20:09
  • 1
    $\begingroup$ The question has changed. $\endgroup$
    – markvs
    Nov 30, 2021 at 20:18
  • 1
    $\begingroup$ 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?) $\endgroup$ Nov 30, 2021 at 20:20
  • $\begingroup$ @BarryCipra But what about $a^2$' why it belongs to $H$ or $Ha$? $\endgroup$
    – Xavi
    Nov 30, 2021 at 20:22
  • $\begingroup$ @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. $\endgroup$ Nov 30, 2021 at 20:24

3 Answers 3

3
$\begingroup$

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 $a \in H$.

$\endgroup$
1
  • $\begingroup$ 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 $a \not \in H$. $\endgroup$
    – Mike
    Nov 30, 2021 at 22:08
3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

$H$ 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 $a^2\in H$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .