# Let $f:G→G'$ a homomorphism and let $K=\ker(f)$ and let $H$ an subgroup of $G$. Prove that: $f^{-1}(f(H))=HK=\{hk:h∈H,k∈K\}$

Let $f:G→G'$ a homomorphism and let $K=\ker(f)$ and let $H$ an subgroup of $G$. Prove that:

$$f^{-1}(f(H))=HK=\{hk:h∈H,k∈K\}$$

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Do your own homework. – Rhys Mar 19 '13 at 12:11

Hint

$$f^{-1}(f(H)) = \{ x \in G : f(x) \in f(H) \} = \{ x \in G : \text{there is h \in H such that f(x) = f(h)} \} = \dots$$

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By the first isomorphism theorem, we know that $G'\cong G/K.$ If we identify $G'$ and $G/K$, then $f(H)$ is the set of cosets $hK$ for $h\in H$. Now, what set in $G$ is mapped surjectively to this set of cosets?

More Deatils. If $g\in G$ is mapped to a coset $hK$, for $h\in H$, then $gK=hK$, that is to say, $g\in hK$, and so the preimage of $f(H)$ under $f$ is the set $HK$.

Inform me of any appearance of errors. Thanks.

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