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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 $a^k$. Tags cardinality, cosets, group-theory, subgroups Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post BenFRayfield Number Theory 0 February 15th, 2014 02:55 PM gaussrelatz Algebra 1 October 10th, 2012 11:30 PM DanielThrice Abstract Algebra 1 November 25th, 2010 02:28 PM sto4432 Abstract Algebra 3 April 30th, 2009 07:17 PM bjh5138 Abstract Algebra 2 August 14th, 2008 12:03 PM

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