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Prove or disprove: if $f:G \rightarrow H$ is a surjective group homomorphism, then

1) $f(A)$ normal implies $A$ normal.

2) $S$ is a p-Sylow subgroup of $G$ implies $f(S)$ is a Sylow-p subgroup of $H$.

I believe 1 is false since I believe we need injectivity, but I cannot think of a counterexample...

I believe 2 is true but do not know how to show it...

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1 Answer 1

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Hints:

For (1) consider a projection $G \times G' \to G'$. What is the image of a subgroup in $G$?

For (2) use the fact that every surjection is of the form $G \to G/H$. Then $S$ goes to $SH/H \simeq S/(S \cap H)$. If $|S| = p^k$ then $|S \cap H| = p^t$ for some $t \leq k$. As $S \cap H$ is also a subgroup of $H$ we get that $p^t \ | \ |H|$ so what is the largest power of $p$ that can divide $|G/H|$?

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A few questions... what is G'? I am not familiar with such notation...also, for 2, doesn't S go to S/H, not SH/H? –  Johnny Apple Aug 24 '13 at 5:21
    
$G'$ is just another group, it's like saying $G_1$. –  Jim Aug 24 '13 at 5:24
    
The notation $S/H$ is generally reserved for when $H \leq S$. The image of $S$ and $SH$ are identical but $H \leq SH$. –  Jim Aug 24 '13 at 5:25
    
The image of a subgroup of H is a singleton element of G' I believe under the projection –  Johnny Apple Aug 24 '13 at 5:26
    
Is that normal? –  Jim Aug 24 '13 at 5:28

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