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My question states, Let H be subgroup of a group G. Prove that HH=H and that $H^{-1}=H$
I started the 1st proof by saying
HH=H
We will show that $HH\subseteq H$
Then $x\in HH$
Then $x=ab$ where $a\in H$ and $b\in H$
I am lost right now trying to prove HH=H and $H^{-1}=H$
Anybody can help me?

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    those are my next few steps for the 1st proof? – 2017-02-11
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    I have not got that far yet because I am still trying to prove HH=H first – 2017-02-11
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    would $H^{-1} \subseteq H$? – 2017-02-11
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    Working on a master's degree in mathematics but not being able to solve the above elementary problem should trigger some serious soul searching about becoming a professional mathematician. – 2017-10-20
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    some professors didnt teach me proofs when i took undergrad courses so i am behind the 8 ball. i learned a lot since. thanks for the concern – 2017-11-13

4 Answers 4

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I already mentioned in my comment that $HH=H$.

We know that $H

Let $x\in H^{-1}$. Then $x=h^{-1}$ for some $h\in H$. Because $H

Let $y\in H$. Then $y^{-1}\in H$. Hence, $y=(y^{-1})^{-1}\in H^{-1}$. Thus, $H\subset H^{-1}$.

Done.Hope it helps.

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    what is the purpose of using Y? – 2017-02-11
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    What? I can't understand. There is no $Y$ in my proof. Rather I used $y$. – 2017-02-11
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    sorry typo. that $y$. what is the purpose of introducing it – 2017-02-11
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    We know that we want to show that $H\subset H^{-1}$. Pick an element in $H$. Its just happened that I denote it by $y$. That's only a name. You can use $x$ if you want.--) – 2017-02-11
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    I did check it. I was the 1st one to do so because I immediately saw that is was right. I just needed a clarification on that $y$ – 2017-02-11
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    I put my answer up. Can you check it? – 2017-02-12
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$\forall a \in H \text{ and } b \in H \implies ab=x \in H$ (because $H$ is a group) $\implies HH \subseteq H$. And $\forall x \in H \implies x = 1.x \in HH \implies H \subseteq HH$ (where $1$ denotes the identity element of $H$).

Now, go ahead with the next one similarly!

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    This look similar to mines up above. It looks like you stated what I did but in reverse order. So basically my next step is for all $x\in H$ if x=1? – 2017-02-11
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    Not really. Lemme try again. Let $1$ be the identity element of $G$, since $H$ is a subgroup, $1 \in H$. So, we have $x = 1.x = x.1 \forall x \in H$. But, $1.x \in HH \implies x \in HH \implies H \subseteq HH$ – 2017-02-11
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Okay here is attempt to answer $HH=H$
We will show that $HH\subseteq H$
Then $x\in HH$
Then $x=ab$ where $a\in H$ and $b\in H$
but $x$ is closed under multiplication
so $x\in H$
My professor also wants me show that $HH\subseteq H$ for my first part of this question. For some reason I do not understand. Somebody can help me?
Here is my attempt to prove $HH\subseteq H$
$x\in H$ if $x\in HH$
I am lost after this

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    Your proof is not properly constructed. – 2017-02-12
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    On your first part, you already said that $x\in HH$ and so $x=ab$. Since $H$ is closed under multiplication, $ab\in H$. So, $x\in H$. With this, we conclude that $HH\subseteq H$. – 2017-02-12
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    ok. my professor told me to show $HH=H$ and $HH\subseteq H$. So I am done then – 2017-02-12
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    On your second part, you need to show that $H\subseteq HH$. Choose an $x\in H$. How do we show that $x\in HH$? Well, we should show that $x$ can be expressed as a product of two elements of $H$. The two elements that we are going to choose are $1$ and $x$, where $1$ is the identity element of group $G$ and hence the identity element of $H$. Now, $$x=1\cdot x.$$ Note that $1\cdot x\in HH$ and this shows that indeed, $x\in HH$. And we are done. Hope this helps. – 2017-02-13
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    so after I state $x\in HH$, state what you just said about the identity element of $G$. cool and thanks again – 2017-02-13
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    got you. thanks again – 2017-02-13
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Given that H be a sub group of G So we have to show that HH = H For that it is enough to show that HH ⊆ H and H ⊆ HH Now let h_1 h_2∈HH where h_1∈ H, h_2∈ H ∴ h_1∈H ,h_2∈ H ⇒ h_1 h_2∈H ( ∵ closed law ) ∴ h_1 h_2 ∈ HH ⇒ h_1 h_2 ∈ H ⇒ HH⊆ H ------------ (1) Again let h ∈ H then we have to show that h ∈ HH Let h ∈ H and e ∈ H ⇒ he ∈ HH ⇒ h ∈ HH

∴ h ∈ H ⇒ h ∈ HH ⇒ H ⊆ HH ------------------------- (2) ∴ from (1) and (2) we get HH = H ( ∵ HH ⊆ H and H ⊆ HH) Hence proved