# Fun with normal subgroups.

• Oct 25th 2010, 10:20 AM
MichaelMath
Fun with normal subgroups.
This stuff drives me bonkers. Anyway, I have been given or have proved the following:

$\displaystyle \phi : X \rightarrow Y$ is onto.

$\displaystyle x, \bar{x} \in X$

$\displaystyle U \subset X$ is a subgroup

$\displaystyle \phi (U)$ is a subgroup of $\displaystyle Y$

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My last question (for this section) is:

Verify that if $\displaystyle U$ is normal in $\displaystyle X$, then $\displaystyle \phi (U)$ is normal in $\displaystyle Y$

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Can I say this:

If $\displaystyle U$ is normal in $\displaystyle X$ then:

$\displaystyle uXu^{-1}=X$ for all $\displaystyle u \in U$ , so:

$\displaystyle \phi(uXu^{-1})= \phi (X)$

$\displaystyle \phi(u)\phi(X) \phi(u^{-1})= \phi (X)$

$\displaystyle \phi(u) Y \phi(u)^{-1}= Y$ for all $\displaystyle \phi(u) \in \phi(U)$

Hence, $\displaystyle \phi(U)$ is normal in $\displaystyle Y$
• Oct 25th 2010, 11:44 AM
Bruno J.
That's perfect. Of course, it all comes down to the fact that $\displaystyle \phi(u)$ can be any element of $\displaystyle Y$. You should perhaps write that down explicitly.