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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 AA = 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 AA 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 AA? 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 AA = {a1a2  a1, a2 in A} ? Because the proof asks for AA. 
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=AA 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  
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra  Quote:
That’s a good question!  
May 20th, 2014, 01:34 AM  #8  
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra  Quote:
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 $a=(mn1)a\in A$. (i) $0\notin A$Let $H=AA$. As $A\ne\emptyset$, i.e. $\exists a\in A$, $0=aa\in H$. Also $H+H=AA+AA=AA=H$. By (i) above $H$ is a subgroup of $G$. Now given $a_1,a_2\in A$, $a_1a_2\in H$ $\implies$ $a_1+H=a_2+H$. 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 $A=A+A\leqslantg+H+A=g+A=A$; hence $A+A=g+H+A$ $\implies$ $A=g+H$, a coset as required.  
May 20th, 2014, 03:06 AM  #9 
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra  

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cardinality, cosets, grouptheory, subgroups 
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