Questions about strong subgroups ($H < G$, $f(H)=H$ for all isomorphisms of $G$)

Question

I need to answer the following question:

Let $G$ be a group. A subgroup $H$ is called strong if for every isomorphism $f$ on G we have $f(H)=H$.

1. Prove: the center of $G$ ($Z(G)=\{x \in G \mid xg=gx, \ \forall g \in G\}$) is a strong subgroup of $G$.
2. Prove that if $H < G$ is strong then $H$ is a normal subgroup of $G$.
3. Prove that if $K$ is a strong subgroup of $H$ and $H$ is a normal subgroup of $G$, then $K$ is a normal subgroup of $G$.

I can't really start with 1. I find this question hard, because I don't really know a way to use $f(H)=H$. I have already noted that isomorphisms on $G$ are in fact permutations. And that the center of $G$ is isomorphic to a subgroup of permutations on $G$, because it is a subgroup of $G$.

Could anybody help me, especially with 1?

Answer 1

So for 1) we need to show that the center is a strong subgroup. So let $x\in Z(G)$ then look at $f(x)\cdot g$ \begin{align*} f(x) g &= f(x)\cdot f(f^{-1}(g)) &\text{Inverse exists because $f$ is an iso}\\ &=f(xf^{-1}(g)) &\text{homomorphism property}\\ &=f(f^{-1}(g)x) & \text{the property of a center}\\ &=gf(x) & \end{align*} and therefore $f(Z(G))\subset Z(G)$. The argument can be repeated with the inverse to show that $f^{-1}(Z(G))\subset Z(G)$ and thus $Z(G)\subset f(Z(G))$. Which proves 1).

2) Now is given that the subgroup is strong. This gives us the liberty to choose the isomorphism. We choose the isomorphism of conjugation $$ f_{g}:G\to G;\; \alpha\mapsto g^{-1}\alpha g $$ Now we use the fact that $H$ is strong to say $f(H)=H$ which translates to $gH=Hg$ so we can conclude that $H$ is a normal subgroup of $G$.

3) Because $H$ is normal in $G$, we have that $f_g(H)=H$ where $f_g$ is the conjugation isomorphism as in 2). In other words this conjugation is an isomorphism on $H$. Thus we have $f_g(K)=K$ since $K$ is strong in $H$. This translates to saying that $K$ is a normal subgroup of $G$.

Answer 2

If $x\in f(Z(G))$, then $xf(g)=f(g)x$ for all $g\in G$. But $f(g)$ runs over all elements of $G$ as well, hence $x\in Z(G)$. So $f(Z(G))\le Z(G)$ and by symmetry $f(Z(G))=Z(G)$.

For 2 and 3 note that conjugations are automorphisms