# Is every normal subgroup the kernel of some endomorphism?

Let $G$ be a group, and let $N$ be a normal subgroup of $G$. Then there is the canonical homomorphism $\phi$ of $G$ onto $G/N$ with kernel $N$. This homomorphism is defined as follows: $\phi(g) \colon= Ng \$ for all $g \in G$.

In which cases can we also define a homomorphism $\psi$ of $G$ into $G$ with kernel equal to $N$? Of course we can do so when $N$ is either $\{e\}$ or $G$ itself. Could there be any other cases when this is possible?

I know that this is not always possible: e.g., consider $Z$ under addition with subgroup $2Z$. Now if there were a homomorphism $\psi$ of $Z$ into $Z$ with kernel equal to $2Z$, then we would have $Z_2 \cong Z/2Z$, which in turn is isomorphic to a subgroup of $Z$, thereby making $Z_2$, a group of order $2$, isomorphic to a subgroup of $Z$. However, every subgroup of $Z$ other than $\{0\}$ is of infinite order.

• You're asking about short exact sequences of groups that split. – Andrew D. Hwang Apr 17 '14 at 14:33

Every normal subgroup $N \leq G$ is the kernel of some homomorphism - namely the canonical map $\pi : G \rightarrow G/N$.

I'm not sure what you mean by the second bit, but if you are denoting by $\mathbb Z_2$ the integers modulo $2$ under addition, then $\mathbb Z/2\mathbb Z$ is isomorphic to $\mathbb Z_2$. There is no contradiction here because $\mathbb Z/2 \mathbb Z$ isn't a subgroup of $\mathbb Z$, but a quotient group.

A restatement of the question is: is every quotient group of a group $$G$$, isomorphic to a subgroup of $$G$$.

As already mentioned, there are many counterexamples.

• the smallest counterexample is the quaternion group $$Q$$ of order 8, which admits the Klein group $$K$$ of order 4 as quotient by its unique subgroup of order 2; since $$K$$ has 3 elements of order 2 and $$Q$$ has only one, $$K$$ is not isomorphic to a subgroup of $$Q$$.
• the second smallest counterexample is the dicyclic group of order 12: indeed its unique subgroup of order 6 is cyclic, while its unique quotient of order 6 is dihedral.
• torsion-free group usually admit non-torsion-free quotients. The simplest example has already been mentioned: $$\mathbf{Z}$$.
• "most" finitely generated groups admits $$2^{\aleph_0}$$ non-isomorphic quotients (necessarily finitely generated), while they have only countably many non-isomorphic finitely generated subgroup. This notably applies to finitely generated non-abelian free groups.
• among torsion abelian groups, here's an example: let $$C_n$$ be cyclic of oder $$n$$; then $$\bigoplus C_{2^n}$$, which is residually finite, has the quotient $$C_{2^\infty}$$, which is not.

Still, there are groups for which every quotient is isomorphic to a subgroup:

• finite abelian groups (more generally, artinian abelian groups)
• simple groups
• finite groups in which every proper quotient is cyclic of prime power order (e.g., simple finite groups, symmetric groups, groups of order $$pq$$ for $$p,q$$ primes)
• finite dihedral groups...

Yes to your question's title [before it was changed as homomorphism $$\to$$ endomorphism] question: in fact, it is a characterization of normal subgroups: a subgroup is normal in its group iff it is the kernel of some homomorphism from that group to another one.

In the case of your "this is not always possible": you're confusing things here. We indeed have that $$\;2\Bbb Z=\ker\phi\;$$ , with $$\;\phi:\Bbb Z\to \Bbb Z_2:=\Bbb Z/2\Bbb Z\;$$ .

It isn't true that $$\;\Bbb Z_2\cong\Bbb Z\;$$ or to a subgroup of it, since any non-trivial subgroup of the integers is isomorphic to the integers, and $$\;\Bbb Z_2\;$$ is finite.

• Unless the title changed after your answer, then your first paragraph is most certainly wrong. He wants an endomorphism, not a homomorphism. – zibadawa timmy Sep 2 '14 at 15:49
• Yes, the title changed. You could have checked this before downvoting... – DonAntonio May 10 '16 at 9:54
• I didn't downvote, just commented. You can edit your answer to accommodate the change in a variety of ways. I don't know why someone with 106k rep and that many badges needs to be told this. Or why it took nearly 2 years for you to care about apparent rep loss. Did you only just recently get downvoted? – zibadawa timmy May 10 '16 at 10:53
• @zibadawatimmy I don't participate anymore in this site, and I got in one week ago or so after almost 10 months without getting in. I really don't care who downvoted, but it's funny that I have some 7,000 points more since the time I left. Weird... – DonAntonio May 10 '16 at 11:02