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For any subgroup of the group $G$, let $H^2$ denote the product $H^2=HH$. Prove that $H^2=H$.

This question seems simple but I do not know how can I prove.

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Hint: Denote the unit element of $G$ as $1_G$. Then $1_G \in H$... –  Johannes Kloos May 8 '12 at 17:03
    
Use that $H$ is closed under multiplication and $1\in H$. | My bad @MattE! –  anon May 8 '12 at 17:04
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@anon: Dear anon, It is stronger than that. Closure under multiplication implies that $H^2 \subset H$, but you need more to get equality. Regards, –  Matt E May 8 '12 at 17:07

3 Answers 3

A start: To prove that two sets $A$ and $B$ are equal, we often use the following strategy: (i) We show that every element of $A$ is an element of $B$ and (ii) We show that every element of $B$ is an element of $A$. It is usually best to tackle these two items separately.

(i) Can you show that everything in $H^2$ is in $H$?

(ii) Can you show that everything in $H$ is in $H^2$?

To do either half, recall the meaning of $H^2$. It is the set of all objects of the form $h_1h_2$, where $h_1\in H$ and $h_2\in H$. The objects $h_1$ and $h_2$ need not be equal. For (ii), the identity element of the group will be useful.

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Its obvious that all elements of $H$ are in $H^2$, right? Think for any $h$ in $H$, 1*$h$ is in $H^2$. On the other hand, for any $h,k$ in $H$, we have $hk$ as an element in $H^2$ (by definition) and as a subgroup is closed under multiplication $hk$ is in $H$. Therefore, $H=H^n$ for all natural numbers.

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First, prove that for all $h\in H$, $Hh=hH=H$.

Now, we are ready to prove that, $HH=H$ :

$$HH=\{h_1h_2|h_1,h_2\in H\}= \underset{h_1\in H}{\cup}\{h_1h_2|h_2\in H\}=\underset{h_1\in H}{\cup}h_1H=\underset{h_1\in H}{\cup}H=H$$

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