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February 27th, 2012, 04:48 PM   #1
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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?
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February 27th, 2012, 05:41 PM   #2
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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.
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February 28th, 2012, 07:10 AM   #3
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Re: on proving a subset is a subgroup

What happens if G is a finite group? Can you still find a counterexample?
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February 29th, 2012, 12:19 PM   #4
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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.
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February 29th, 2012, 12:36 PM   #5
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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?
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March 5th, 2012, 03:35 PM   #6
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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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