# Having all subgroups normal is isomorphism invariant

• Mr Davis 97
In summary, being Hamiltonian is an isomorphism invariant, as proven by showing that if all subgroups of a group are normal and the group is isomorphic to another group, then all subgroups of the other group are also normal. This is established through two lemmas, one showing that the preimage of a subgroup is also a subgroup and the other showing that if the preimage of a subgroup is normal, then the subgroup itself is normal.

## 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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In this case you really do merely need two lines:
Mr Davis 97 said:
##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