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I am reading up on Fraleigh's A First Course in Abstract Algebra, and he says ($H$ subgroup of $G$) $Hg=gH$ $iff$ $i_g[H]=H$ $iff$ $H$ is invariant under all inner automorphisms. I look up invariant and I find this definition:

"Firstly, if one has a group G acting on a mathematical object (or set of objects) X, then one may ask which points x are unchanged, "invariant" under the group action, or under an element g of the group." from Invariant Description Wiki.

First I am wondering if that means the elements of $H$ do not change but change positions (hence the permutation) or is $H$ the identity under all inner automorphisms of $G$. W

EDIT: Too many questions asked by me, I will ask them separately.

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Typical definition: for a set $X$ and a bijection $f:X\to X$ a subset $A\subseteq X$ is invariant under $f$ iff $f(A)=A$. This does not mean that $f$ restricted to $A$ is the identity function.

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  • $\begingroup$ Fantastic! Thanks for clearing that up $\endgroup$ – Kam Dec 7 '17 at 19:00
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Inner automorphisms need not act trivially (fix every element) of $N$ for it to be normal.

This is all just a fancy way of saying that $gNg^{-1}=N$. This precisely what it means to be invariant, it means that applying an inner automorphism to $N$ sends you back to $N$, so that the image of $N$ under conjugation of any element is again $N$.

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    $\begingroup$ This is an answer to your first question, the second question is really $10$ quesitons. I recommend asking different questions, or perusing different answers on SE. $\endgroup$ – Andres Mejia Dec 7 '17 at 18:10
  • $\begingroup$ Thank you for your time! $\endgroup$ – Kam Dec 7 '17 at 18:50
  • $\begingroup$ Just to be clear though, invariant doesn't mean that the elements in N cannot permute, just that their values don't change? $\endgroup$ – Kam Dec 7 '17 at 18:51
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    $\begingroup$ I’m not sure what you mean. Elements don’t have to be fixed at all, their images just also need to lay in the normal subgroup $\endgroup$ – Andres Mejia Dec 7 '17 at 19:10

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