Well some of these groups will be infinite so that's impossible but for finite groups I guess it would work but would be a tad tedious. We're looking for a more generic proof.
Consider the following,
Let H \subset G. The group G is abelian and therefore has commutivity of elements by design i.e.
ah=ha
However, this holds \forall h \in H and \forall a \in G
\Rightarrow aH=Ha \Rightarrow a^{-1}aH=a^{-1}Ha
a^{-1}Ha=H
#5
futurebird
270
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But my group isn't abelian.
#6
latentcorpse
1,411
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haha i am being silly...let me reconsider
#7
latentcorpse
1,411
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could u post the question?
#8
futurebird
270
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It's for a take-home final so I'm trying to ask for help on the concepts without doing that. I'll post it after I turn it in.