If N is characteristic in a group G and if we have that K/N is characteristic in G/N why do we have that K is characteristic in G.

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Since $N$ is characteristic, any automorphism of $G$ descends to $G/N$. Now if $K$ was not characteristic in $G$, what would happen to $K/N$? –  user641 Mar 17 '11 at 2:18

Let $\varphi\colon G\to G$ be a homomorphism of $G$. Consider $K$ and $\varphi(K)$; since $N\subseteq K$, then $N=\varphi(N)\subseteq\varphi(K)$, so then by the lattice isomorphism theorem, $K$ and $\varphi(K)$ correspond to subgroups of $G/N$, and $K=\varphi(K)$ if and only if $K/N = \varphi(K)/N$.

But since $\varphi(N)=N$, then $\varphi$ induces an automorphism $\overline{\varphi}$ on $G/N$ by $\overline{\varphi}(gN) = \varphi(g)N$. This is well defined, since $xN=yN$ if and only if $y^{-1}x\in N$, if and only if $\varphi(y)^{-1}\varphi(x)\in\varphi(N)=N$, so $\varphi(x)N = \varphi(y)N$.

Therefore, $\varphi(K)/N = \overline{\varphi}(K/N) = K/N$ (since $K/N$ is characteristic in $G/N$), so $K=\varphi(K)$. Thus, $K$ is characteristic in $G$.

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This is false. $N/N$ is always characteristic in $G/N$, so if this were true, then normal subgroups would always be characteristic.