Let $H\le G$ s.t. $x^2\in H$ for all $x\in G$. Prove that $H\unlhd G$ and $G/H$ is abelian.

Question

Let $G$ be a group & $H$ be a subgroup of $G$, where $G$ has the property that $x^2 \in H$ for every $x \in G$. Proof that $H$ is normal in $G$ and $G/H$ is abelian.

My approach:

It can be show that $H$ is normal in $G$. Since $H$ is normal in $G$ and $x \in G$, $x^2 \in H$ therefore $[G:H]=2$ (is it logical enough?). Then the coset of $H$ in $G$ are $H$ and $G-H$. If $ x \in G-H$, then $xH=G-H$ and therefore $(G-H)^2=H$, $H$ being the identy element of $G-H$. Hence $G-H$ is a group of prime order hence abelian.

Need a little hint how to show $G/H$ is abelian.

Here is my second attempt:

$H$ is normal in $G$

Let $C$ be the commutator of $a^2$, $b^2$ in $G$.

Therefore $$ a^2b^2(a^2)^{-1}(b^2)^{-1}\in G$$ And therefore $ a^2b^2(a^2)^{-1}(b^2)^{-1} \in H$. Therefore $C$ is a subset of $H$

Then $a^2b^2(a^2)^{-1}(b^2)^{-1} \in H$ i.e. $a^2b^2(b^2a^2)^{-1} \in H$ Therefore $H(a^2b^2)=H(b^2a^2)$ which implies $(Ha^2)(Hb^2)=(Hb^2)(Ha^2)$ And this follows.

My head is not working anymore :(

Compare with this duplicate. DonAntonio shows that $G/H$ is abelian. — Dietrich Burde, Nov 21 '19 at 16:15
It is not true that $[G:H]=2$. Take $\Bbb Z/2\times \Bbb Z/2$ and the trivial group — pancini, Nov 21 '19 at 16:16
I don't understand the downvote. This is an honest question by someone who is clearly trying. — user1729, Nov 21 '19 at 16:25
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. — Shaun, Nov 21 '19 at 16:38
Comment on attempt 2: Don't consider $a^2$ and $b^2$. Instead define $c=ab$ and consider $c^2$. — user1729, Nov 21 '19 at 19:25

score 2 · Accepted Answer · answered Nov 21 '19 at 16:19

If $x^2\in H$ for all $x\in G$ this implies that every element of the quotient group $G/H$ has order two. It does not imply that $[G:H]=2$.

As a counter-example, take $G$ to be the Klein $4$-group and $H$ to be the trivial group.

However, you are nearly there as it is. As you have proven that $H$ is normal in $G$, you just need to prove that if $K$ is a group such that $k^2=1$ for all $k\in K$ then $K$ is abelian. This is a standard exercise, and the idea is: for all $a, b\in K$ we have $$1=(ab)^2=abab=a^{-1}b^{-1}ab.$$ So $a^{-1}b^{-1}ab=1$, which rearranges to give $ab=ba$ as required.

Hmm. I think i got it. Thanks. Let me try it. – Jitu Biswas Nov 21 '19 at 16:31 — Jitu Biswas, Nov 21 '19 at 16:31

score 1 · Answer 2 · answered Nov 21 '19 at 16:21

1

The first part has been proved already, see the duplicate. $G/H$ is abelian, because every element has order $2$ except for the identity. For the proof see the duplicate here:

Order of nontrivial elements is 2 implies Abelian group

answered Nov 21 '19 at 16:21

Dietrich Burde

130,978

score 1 · Answer 3 · answered Apr 07 '23 at 05:43

Let $aH, bH\in G/H$ for any $a,b\in G$. Therefore $a^2,b^2\in H$. Therefore $a^2H,b^2H =H$, so that $aH=a^{-1}H$ and $bH=b{^-1}H$. Therefore $(aH)(bH)=(a^{-1}H)(b^{-1}H)$. Therefore $abH=a^{-1}b^{-1}H=(ba)^{-1}H$. But for any $a\in G, aH=a^{-1}H$ . Therefore $(ba)^{-1}H=baH$. Therefore $abH=baH$. Therefore $(aH)(bH)=(bH)(aH)$. Hence $G/H$ is abelian

Let $H\le G$ s.t. $x^2\in H$ for all $x\in G$. Prove that $H\unlhd G$ and $G/H$ is abelian.

My approach:

Here is my second attempt:

3 Answers3