A subgroup containing a kernel of a group homomorphism into an abelian group is a normal subgroup.

Let $f \colon G \rightarrow H$ be a group homomorphism, where $H$ is an abelian group.

If $\ker f \subset N$ for some $N \subset G$, then $N$ is a normal subgroup of $G$.

I don't know how to start to prove the statement. Any hint?

• $gng^{-1}=(gng^{-1}n^{-1})n$. Jun 12, 2013 at 15:43

By the Lattice Theorem, subgroups of $G$ containing $\ker f$ are in one-to-one correspondence with subgroups of $G/\ker f$, and by the first isomorphism theorem,

$$G/\ker f \cong f(G)$$

So all subgroups of $G/\ker f$ are normal, and the corresponding subgroups are normal .

Hint: Since $H$ is abelian, $a^{-1}b^{-1}ab \in \operatorname{Ker}(f)$ for all $a, b \in G$.

Hint: Every subgroup of an abelian group is normal.

• But $G$ is not abelian here. What am I missing? Jun 12, 2013 at 15:39
• @atricolf the isomorphism theorems :-) subgroups containing the kernel are the same thing as subgroups of the image Jun 12, 2013 at 15:44

HINT n.3 given $\phi$ homomorphism $G \to H$ for every $N$ normal in $H$, $\phi^{-1}(N)$ is normal in $G$. Combine this with the other hints and the fact that for every $A \leq G$, $\phi(A) \leq H$ and you win :)

let $f:G_{1}\rightarrow~G_{2}$ be the group homomorphism, where $G_{2}$ is abelian.\ let $H_{1}$ be the subgroup of $G_{1}$ s.t. $ker(f)\subseteq~H_{1}$\ $f(h_{1}^{-1}g_{1}h_{1}g_{1}^{-1})=e_{2}$, where $g_{1}\in~G_{1}$ and $h_{1}\in~H_{1}$ and $e_{2}$ is the identity of $G_{2}$\ Since $G_{2}$ is abelian and $ker(f)\subseteq~H_{1}$ so we can easily have $g_{1}h_{1}g_{1}^{-1}\in~H_{1}$\ Hence $H_{1}$ is normal.

$$f:G\rightarrow~H$$, by the first isomorphism theorem,

$$G/\ker f \cong f(G)$$

So you need to create a correspondance $$G/N \rightarrow f(G)$$ ,

now $$f^{-1}(f(h'h))=f^{-1}(f(hh'))$$ $$f^{-1}(f(h')f(h))=f^{-1}(f(h)f(h'))$$then $$gN=Ng$$ if ker(f) $$\subset$$ N, you can map element $$e^H$$ to ker(f) inside N for all elements.