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Let $G$ be an abelian group. Prove that $H = \{a \in G \mid a^2 = e\}$ is subgroup of $G$, where $e$ is the neutral element of $G$.

I need some help to approach this question.

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marked as duplicate by Cameron Buie, Asaf Karagila, Tom Oldfield, Amzoti, vadim123 May 18 '13 at 19:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

What is the definition of a subgroup? What have you previously done to prove something is a subgroup? – Erick Wong May 18 '13 at 18:27
Sub group is a sub set in G with meets the group axioms – Billie May 18 '13 at 18:29
@user1798362: Can you state those axioms? Which ones do you not know how to prove that this $H$ satisfies? – Zev Chonoles May 18 '13 at 18:30
up vote 2 down vote accepted

Hint: To show it is a subgroup you must show the following three things:

$1)$ $e \in H$.

$2)$ If $a$ and $b$ are in $H$ then so is $ab$.

$3)$ If $a$ is in $H$, then so is $a^{-1}$.

To show these three things, think about what it means for a particular element to be in $H$ and try and use this to show that the elements you want to be in $H$ actually are.

Once you have shown this is true, you might like to think about why we needed the group to be abelian, and see if you can find a (non-abelian) group where $H$ is not a subgroup.

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Hint: To show it is a subgroup, you need to answer yes to the following questions, and justify your answers.

Questions to answer:

$(1)$ What is $e^2\,$? So is $\,e \in H\,$?

  • (This will prove the identity of $G$ is in $H$.)

$(2)$ If $\,a, b \in H$, then $a^2 = b^2 = e.\;$ So we need to know whether $(ab) \in H$?:
$Yes, ab \in H$ if you can show that $(ab)^2 = e$.
Tip: use the fact that $G$ is abelian, and so $(ab)^2 = a^2b^2$

  • (If $ab\in H$, this would prove that $H$ is closed under the group operation of $G$).

$(3)$ If $a \in H$, is $a^{-1} \in H$?

  • (If so, this would prove that $H$ is closed under taking inverses.)
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Very nice hint and guidance! +1 – Amzoti May 19 '13 at 0:43

Hint: Find a homomorphism from $G$ whose kernel is exactly $H$.

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Prove that if $\,G\,$ is an abelian group then $\,G^n:=\{ g^n\;;\;g\in G\}\,$ is a subgroup of G .

Further hint: $\,(xy)^n=x^ny^n\,$ ...what about inverses?

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If $a,b \in H $ then $abab=abba=ab^2a=aea=a^2=e \in H$


If $a,b,c \in H$ then $(ab)c=a(bc)$ because $G$ is a group


$e\in H$ because $ee=e^2=e$


Let $a \in H$ Then $a^2=e=aa^{-1}=a^{-1}a$ Multiply by $a$ then $a^{-1}aa=aa^{-1}a \rightarrow a^{-1}e=ea \rightarrow a^{-1}=a \in H$. The object $a^{-1}\in G$ exists because $G$ is a group

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