# Having all subgroups normal is isomorphism invariant

## Homework Statement

A group is called Hamiltonian if every subgroup of the group is a normal subgroup. Prove that being Hamiltonian is an isomorphism invariant.

## The Attempt at a Solution

Let $f$ be an isomorphism from $G$ to $H$ and let $N \le H$. First we prove two lemmas:

Lemma 1): If $f^{-1}(N) \le G$ then $N \le H$.

Let $a,b \in N$. Then since $f$ is surjective, $a=f(x)$ and $b=f(y)$ for some $x,y \in f^{-1}(N)$. Then $ab^{-1} = f(x)f(y)^{-1} = f(xy^{-1}) \implies xy^{-1} = f^{-1}(ab^{-1})$. In the previous step we used the fact that $f$ is a homomorphism and that $f$ is injective. Now, since $x,y \in f^{-1}(N)$ and $f^{-1}(N)$ is a subgroup, we have that $xy^{-1} \in f^{-1}(N)$. So $f^{-1}(ab^{-1} \in f^{-1}(N)$. By definition of the preimage, this means that $f(f^{-1}(ab^{-1})) \in N \implies ab^{-1} \in N$. So $N \le H$.

Lemma 2): If $f^{-1}(N) \trianglelefteq G$ then $N \trianglelefteq H$.

Here we use lemma 2 and suppose that $N \le H$. We want to show that $\forall h \in H$, $hNh^{-1} \subseteq N$. So, let $h\in H$ and $n \in N$. Since $f$ is surjective, $h = f(y)$ for some $y \in G$ and $n=f(x)$ for some $x \in f^{-1}(N)$. Then $hnh^{-1} = f(y)f(x)f(y)^{-1} = f(xyx^{-1}) \implies yxy^{-1} = f^{-1}(hnh^{-1})$. But $f^{-1}(N) \trianglelefteq G$, so $yxy^{-1} \in f^{-1}(N)$. Hence $f^{-1}(hnh^{-1}) \in f^{-1}(N)$. By definition of the preimage, this implies that $hnh^{-1} \in N$. We conclude that $N \trianglelefteq H$.

Finally, we prove the main result, that if all of the subgroups of $G$ are normal, and if $G$ is isomorphic to $H$, then all of the subgroups of $H$ are normal.

Let $N$ be a subgroup of $H$. We know that the preimage of a subgroup is also a subgroup, so $f^{-1}(N) \le G$. But all of the subgroups of $G$ are normal, so $f^{-1}(N) \trianglelefteq G$. By lemma 2, we then know that $N \trianglelefteq H$, and we are done.

NOTE: Perhaps this argument would have been easier if I supposed that $f : H \to G$ were the isomorphism.

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fresh_42
Mentor
In this case you really do merely need two lines:
$xy^{-1} \in f^{-1}(N)$. So $f^{-1}(ab^{-1} \in f^{-1}(N)$.
and
Then $hnh^{-1} = f(y)f(x)f(y)^{-1} = f(xyx^{-1}) \implies yxy^{-1} = f^{-1}(hnh^{-1})$.
Let me write it a bit easier. Say $f\, : \,G \longrightarrow H$ is an isomorphism and $N \trianglelefteq G$ and $n,m \in N, g \in G, h=f(g)\in H\,.$

Then $f(n)f(m)^{-1}=f(nm^{-1}) \in f(N)$ shows the subgroup property, and $hf(N)h^{-1}=f(gNg^{-1}) \subseteq f(N)$ shows normality. That's all.

• Mr Davis 97