Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
@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

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.

share|cite|improve this answer

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.

share|cite|improve this answer

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

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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.