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quasar987

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Is this correct? The wiki article seems to indicate that equality btw xHx^{-1} and H is unnecessary, but rather that inclusion is sufficient.

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- #1

quasar987

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Is this correct? The wiki article seems to indicate that equality btw xHx^{-1} and H is unnecessary, but rather that inclusion is sufficient.

- #2

matt grime

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They are easily shown to be equivalent.

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quasar987

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Mmmh... So for all x in G, [itex]xHx^{-1}\subset H \Leftrightarrow xHx^{-1}= H[/itex].

For a fixed x, suppose there exists h' in H such that there are no h in H with h'=xhx^{-1}. This is nonsense since h=x^{-1}h'x does the trick.

For a fixed x, suppose there exists h' in H such that there are no h in H with h'=xhx^{-1}. This is nonsense since h=x^{-1}h'x does the trick.

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- #4

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But then how do you know h is in H? And I don't see what this has to do with your original question.

- #5

quasar987

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My original question was (essentially) *Are the two definitions of normal subgroup "H is is normal subgroup of G if for all x in G, [itex]xHx^{-1}\subset H[/itex]" and "H is is normal subgroup of G if for all x in G, [itex]xHx^{-1}= H[/itex]" equivalent?*

Then matt grime said "Yes". Then I tried to prove this assertion, so I said the following: "To show they are equivalent, I only need to show that for all x in G, [itex]xHx^{-1}\subset H \Rightarrow xHx^{-1}= H[/itex] since the converse is immediate."

So let's proceed by contradiction by supposing that for a certain fixed x in G, there exists an h' in H such that there are no h in H with h' = xhx^{-1}. This is automatically seen to be an absurdity since the element h of G defined by h=x^{-1}h'x is such that h' = xhx^{-1} and it is in H because by hypothesis, h' is in H and [itex]xHx^{-1}\subset H[/itex], i.e. [itex]h=x^{-1}h'x\in H[/itex].

I had to write it all in large to convince myself fully that it is correct, and still seems so to me. Do you still have an objection StatusX?

Then matt grime said "Yes". Then I tried to prove this assertion, so I said the following: "To show they are equivalent, I only need to show that for all x in G, [itex]xHx^{-1}\subset H \Rightarrow xHx^{-1}= H[/itex] since the converse is immediate."

So let's proceed by contradiction by supposing that for a certain fixed x in G, there exists an h' in H such that there are no h in H with h' = xhx^{-1}. This is automatically seen to be an absurdity since the element h of G defined by h=x^{-1}h'x is such that h' = xhx^{-1} and it is in H because by hypothesis, h' is in H and [itex]xHx^{-1}\subset H[/itex], i.e. [itex]h=x^{-1}h'x\in H[/itex].

I had to write it all in large to convince myself fully that it is correct, and still seems so to me. Do you still have an objection StatusX?

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- #6

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Assume for all x in G, xHx^-1 is a subset of H. Then for all h in H, we can define h'=x^-1hx, which is in H, and so h=xh'x^-1, ie, H is a subset of xHx^-1.

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