# Problem of the Week #24 - November 12th, 2012

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#### Chris L T521

##### Well-known member
Staff member
Here's this week's problem.

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Problem: Let $G$ be a group and let $f:G\rightarrow H$ be a group homomorphism. If $U\leq G$, show that $f^{-1}(f(U))=U\ker(f)$.

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Note: In this case $f^{-1}$ refers to the pre-image of $f$, not it's inverse!

#### Chris L T521

##### Well-known member
Staff member
This week's question was correctly answered by Deveno. His answer can be found below.

we will show the two sets are equal by showing they contain each other.

suppose that $$\displaystyle g \in Uker( f)$$.

then $$\displaystyle g = uk$$ for some $$\displaystyle u \in U, k \in ker( f)$$.

hence $$\displaystyle f(g) = f(uk) = f(u)f(k) = f(u)e_H = f(u)$$ (because f is a homomorphism), so $$\displaystyle f(g) \in f(U)$$,

so $$\displaystyle g \in f^{-1}(f(U))$$.

on the other hand, suppose $$\displaystyle g \in f^{-1}(f(U))$$.

this means that $$\displaystyle f(g) \in f(U)$$, so $$\displaystyle f(g) = f(u)$$ for some $$\displaystyle u \in U$$.

hence $$\displaystyle (f(u))^{-1}f(g) = e_H$$, and because f is a homomorphism:

$$\displaystyle (f(u))^{-1}f(g) = f(u^{-1})f(g) = f(u^{-1}g)$$,

so $$\displaystyle f(u^{-1}g) = e \implies u^{-1}g \in ker(f)$$,

so $$\displaystyle u^{-1}g = k$$, for some $$\displaystyle k \in ker(f)$$,

thus $$\displaystyle g = uk \in Uker(f)$$, QED.

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