# If two quotients groups of finitely generated group are isomorphic, then corresponding normal subgroups are isomorphic too?

In Topics in Geometric Group Theory, section 3.C lemma 42,

For a finitely-generated group T, the following two properties are equivalent:

(i) T has uncountably many normal subgroups,

(ii) T has uncountably many non-isomorphic quotients.

In proof, it assumes that the cardinality of the set of normal subgroup of T is uncountable and the cardinality of the set of isomorphism classes of quotient groups of T is countable to obtain a contradiction. it shows that there exist an uncountably many homomorphisms from T onto quotient group Q. but the cardinality of the set of Q is countable as above assumption, it is contradiction.

Here is my question. In the last part of the proof, it has to be proven first that the group homomorphisms are injective, or If two quotients groups of finitely generated group are isomorphic, then corresponding normal subgroups are isomorphic. is it correct? if not, tell me why it is contradiction specifically please.

• The main idea for this equivalence is that for a finitely generated group $G$, the equivalence relation on the set of normal subgroups of $G$: [$N\sim P$ if $G/N$ and $G/P$ are isomorphic] has countable classes. – YCor Dec 30 '17 at 6:28

## 2 Answers

This isn't true even for finite groups. Let $D_4$ be the dihedral group of order $8$. Then $D_4$ has subgroups of order $4$ isomorphic to $C_4$ (cyclic order $4$) and $V_4$ (Klein four-group). They are both normal and the quotients are both cyclic of order $2$.

• Then can you give me the reason why this is contradiction? if the group homomorphisms are not injective, it can not be said contradiction, i think.. – mwoarhkd Dec 29 '17 at 13:27

You write "it has to be proven first that the group homomorphisms are injective", but you don't say which group homomorphisms.

Here is a direct proof (i.e. a proof without contradiction) that has no injective group homomorphisms.

Consider a countable set $A$ of quotient groups of $T$, one for each isomorphism class. For each $Q \in A$ consider the countable set $H_Q$ of homomorphisms $h : T \to Q$. The set $$H = \bigcup_{Q \in A} H_Q$$ is countable. By taking the kernel you get a surjection from the countable set $H$ onto the set of normal subgroups of $T$, and so the latter set is countable.