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February 27th, 2012, 04:48 PM  #1 
Senior Member Joined: Feb 2012 Posts: 628 Thanks: 1  on proving a subset is a subgroup
If I define a subset of a group G = H by the characteristics that the elements of H have in common, the identity element is in H, and I show that the subset is closed under the group operation, must the subset H be a subgroup of G? Or is there a way for H not to be a subgroup even though all of these things are true about it?

February 27th, 2012, 05:41 PM  #2 
Global Moderator Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4  Re: on proving a subset is a subgroup
Like the set of positive integers, with zero? They all have "being positive" in common, and are closed under addition. But this is not a subgroup of the integers with addition. A necessary and sufficient condition for being a subgroup is that for all x, y in the subset, xy^1 is in the subset. Unless I'm wrong... !!! Sadly, I have to put such disclaimers when I type from work on my phone. 
February 28th, 2012, 07:10 AM  #3 
Senior Member Joined: Feb 2012 Posts: 628 Thanks: 1  Re: on proving a subset is a subgroup
What happens if G is a finite group? Can you still find a counterexample?

February 29th, 2012, 12:19 PM  #4 
Newbie Joined: Feb 2012 Posts: 17 Thanks: 0  Re: on proving a subset is a subgroup
If every element of your subset has a finite order, closure under the group operation will give you the inverses you need.

February 29th, 2012, 12:36 PM  #5 
Senior Member Joined: Feb 2012 Posts: 628 Thanks: 1  Re: on proving a subset is a subgroup
So basically if G is a finite group and I can prove that the subset H is closed under the group operation, then the only thing left to show is that the identity element is in H and then I will have proven that H is a subgroup of G?

March 5th, 2012, 03:35 PM  #6 
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88  Re: on proving a subset is a subgroup
for a finite subset of a group (even if the group is infinite), closure is sufficient: suppose H is a finite subset of a group, closed under multiplication. let h be any element of H. then the map x>hx is a set bijection of H with itself, since if hx = hy, then (in G) we have h^1(hx) = h^1(hy), so x = y. since H is finite, any injection H>H is bijective. thus there is some element of H, say z, with hz = h (bijective implies onto, and h is certainly in H). but e in G is the unique element of G with this property, hence z = e, so e is in H. moreover, we then have, for each h, an element w in H with hw = e (since e is in H and the map x>hx is onto). but this shows that h^1 also lies in H, so H is a subgroup of G. 

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