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This is probably really basic, but here it goes

Let K be a subgroup of some group H; let X be the set of left cosets of K, i.e. X={hK:h∈H}; and let G be the group of permutations of X. For all h∈H, let f(h)∈G be the permutation of X that sends every coset h′K to the coset hh′K.

Now I know that Ker(f)={h∈H:haK=aK for every a∈H}=⋂a∈Ha−1Ka is a normal subgroup of H. What I don't understand is why (as my text claims) Ker(f) is contained in K.

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Observe $k\in\ker f\implies kK=K \implies k\in K$.

(Also notice $K$ is one of the sets you're intersecting in $\displaystyle\bigcap_{a\in H}aKa^{-1}$.)

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  • $\begingroup$ Thank you. The insight about intersection makes sense. As for the first line though wouldn't you have k ∈ ker f ⟹ kaK = aK ? $\endgroup$ – Francis H White Sep 22 '15 at 4:14
  • $\begingroup$ @FrancisHWhite You have $kaK=aK$ for all $a$. All $a$. That means in particular we have $kK=K$. $\endgroup$ – whacka Sep 22 '15 at 4:16
  • $\begingroup$ very helpful. thanks $\endgroup$ – Francis H White Sep 22 '15 at 4:18
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$\ker(f)=\{h\in H : a^{-1}ha\in K\ \forall a \in H\}$ implies that $\ker$ is contained in $K$ as if you will take $a=e$ (as you see it is for all $a\in H$), it clearly implies that any element of $\ker$ lies in $K$ (as $hK=K \implies h \in K$).

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