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Dummit and Foote leave this as an "easy exercise for verification" and I would like to verify that I am indeed reasoning correctly.

Proof Take an arbitrary element $kH \in K/H$, and let $G/H$ act on $K/H$ by conjugation. $$(gH)(kH)(gH)^{-1} = (gH)(kH)(H^{-1}g^{-1}) = gh_1 kh_2 h_3^{-1} g^{-1} = gh_1kh_4g^{-1}$$ But $H\le K$, and so $h_1 \in H$ is also in $K$, then $h_1 kh_4 \in K$, and $K$ is normal in $G$, so $gKg^{-1} = K$, and thus $(gH)(kH)(gH)^{-1} \in K$.

My Question

How do I recover my coset $kH \in K/H$? I have made an error along the way, and can't see where my error lies.

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The product of elements in $G/H$ is given by $(g_1H)(g_2H) = g_1g_2H$. I'm not sure what you're doing exactly. –  Cocopuffs Nov 3 '12 at 15:44
    
@Cocopuffs jmi4's proof is not efficient, but it is nonetheless correct. –  Phira Nov 3 '12 at 15:46
    
@Phira I see, I was confused by the switch between left cosets and right cosets or elements.. –  Cocopuffs Nov 3 '12 at 15:55

1 Answer 1

up vote 4 down vote accepted

The second equality sign in your proof is not an equality, as you go from a set notation to a notation that gives a generic member of the set. At the end, you did the reverse notational misstep, by writing a set being an element instead of a subset of another set.

So, you start out with a set that you want to be a coset of type $k'H$ with $k' \in K$, (not necessarily the same $k$!) so it is enough to show that your generic element can be written as $k'h'$ with $k' \in K$ and $h' \in H$, but since $H$ is a subgroup of $K$ this is exactly the same thing as showing that the generic element is in $K$ which is precisely what you have done (up to notational problems).


For completeness, let me detail Cocopuffs' comment:

Since $H$ is normal, you can calculate with the cosets "the easy way":

$(gH)(kH)(gH)^{-1} = (gkg^{-1})H= k'H$ where the first equality uses the normality of $H$ and the second equality uses the normality of $K$-

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Understood. Thanks for the precise explanation –  Zvpunry Nov 3 '12 at 18:52

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