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 April 2nd, 2014, 10:39 AM #1 Newbie   Joined: Mar 2014 From: Canada Posts: 5 Thanks: 0 Subgroups, Cardinality Let (G,+) be an abelian group and let A be a finite subset of G. Prove that |A-A| = |A| iff A is a coset of a subgroup of G. I'm not sure how to go about this ..
 April 2nd, 2014, 12:58 PM #2 Senior Member     Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra Could you explain what your notation |A-A| means? I don’t think I’ve seen it before.
 April 2nd, 2014, 01:10 PM #3 Newbie   Joined: Mar 2014 From: Canada Posts: 5 Thanks: 0 It's the number of elements in the set.
 April 2nd, 2014, 02:06 PM #4 Senior Member     Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra Yes, what do you mean by A-A? I think you want $$A+A\ =\ \{a_1+a_2:a_1,a_2\in A\}$$ Now $A$ is a coset, i.e. $A=g+H$ for some $g\in G$ and subgroup $H$. Then $$A+A$$ $=\ \{g+h_1+g+h_2:h_1,h_2\in H\}$ $=\ \{2g+h_1+h_2:h_1,h_2\in H\}$ (since $G$ is Abelian) $=\ \{2g+h:h\in H\}$ $=\ 2g+H$ But $2g+H$ is another coset and so has the same number of elements as $g+H=A$.
 April 2nd, 2014, 03:50 PM #5 Newbie   Joined: Mar 2014 From: Canada Posts: 5 Thanks: 0 Along, the same lines, wouldn't A-A = {a1-a2 | a1, a2 in A} ? Because the proof asks for |A-A|.
 April 2nd, 2014, 03:57 PM #6 Newbie   Joined: Mar 2014 From: Canada Posts: 5 Thanks: 0 But to work off your proof (since I have to prove it for A+A also), how would I prove it the other way? So assuming |A+A| = |A|, how do we show A is a coset of a subgroup of G. I think I have to use H=A-A as the subgroup but not sure about the technique of the proof Last edited by orangestripes; April 2nd, 2014 at 04:18 PM.
April 2nd, 2014, 04:19 PM   #7
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Quote:
 Originally Posted by orangestripes Along, the same lines, wouldn't A-A = {a1-a2 | a1, a2 in A} ? Because the proof asks for |A-A|.
In that case the problem would be more or less the same as $\displaystyle A-A$ is just the subgroup $\displaystyle H$ itself.

Quote:
 Originally Posted by orangestripes But to work off your proof (since I have to prove it for A+A also), how would I prove it the other way? So assuming |A+A| = |A|, how do we show A is a coset of a subgroup of G.
That’s a good question!

May 20th, 2014, 01:34 AM   #8
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Quote:
 Originally Posted by orangestripes But to work off your proof (since I have to prove it for A+A also), how would I prove it the other way? So assuming |A+A| = |A|, how do we show A is a coset of a subgroup of G. I think I have to use H=A-A as the subgroup but not sure about the technique of the proof
After chewing over this problem for nearly two months, I think I’ve finally got the solution.

We split it into two cases: (i) $0\in A$, and (ii) $0\notin A$.
(i) $0\in A$
We shall show that in this case $A$ is subgroup (and hence coset).

We have $\forall a\in A$ $a=a+0\in A+A$ $\implies$ $A\subseteq A+A$. Hence, since $A$ and $A+A$ are finite with the same number elements, $A=A+A$. This means that $A$ is closed under $+$.

It remains to show that $-a\in A$ for all $a\in A$. By closure $a^k\in A$ for all integers $k\geqslant0$. But $A$ is finite so $\exists m,n$ with $m>n$ such that $ma=na$. Hence $-a=(m-n-1)a\in A$.
(i) $0\notin A$
Let $H=A-A$. As $A\ne\emptyset$, i.e. $\exists a\in A$, $0=a-a\in H$. Also $H+H=A-A+A-A=A-A=H$. By (i) above $H$ is a subgroup of $G$.

Now given $a_1,a_2\in A$, $a_1-a_2\in H$ $\implies$ $a_1+H=a_2+H$. Hence any two elements in $A$ belong to the same coset of $H$. Thus $A$ is a subset of a coset of $H$, say $A\subseteq g+H$ for some $g\in G$. Then $|A|=|A+A|\leqslant|g+H+A|=|g+A|=|A|$; hence $A+A=g+H+A$ $\implies$ $A=g+H$, a coset as required.

May 20th, 2014, 03:06 AM   #9
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Sorry …
Quote:
 Originally Posted by Olinguito $a^k\in A$
That should be $ka$, not $a^k$.

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