# 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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## Answers and Replies

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