Let be a group and let be a subgroup of . Using the definitions of the sets , , and , prove that for every in if and only if . = { in | }
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Originally Posted by ll2008 Let be a group and let be a subgroup of . Using the definitions of the sets , , and , prove that for every in if and only if . = { in | } if and only if is normal subgroup if and only if the normalizer of is .
I thought about that, but I don't think I can use those theorems to prove it. I would have to prove it using the definitions.
Hacker's way is valid. It's using the definitions, not theorems. Here it is essentially broken down into baby steps. If , then implies , whence . If for all , it follows that (and so equality too of course).
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