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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.

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    $\begingroup$ You're asking about short exact sequences of groups that split. $\endgroup$ Apr 17, 2014 at 14:33

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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 groups of order $\prod_i p_i$ with $p_i$ distinct primes (every quotient map then splits)
  • finite dihedral groups...
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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.

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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.

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    $\begingroup$ Unless the title changed after your answer, then your first paragraph is most certainly wrong. He wants an endomorphism, not a homomorphism. $\endgroup$ Sep 2, 2014 at 15:49
  • $\begingroup$ Yes, the title changed. You could have checked this before downvoting... $\endgroup$
    – DonAntonio
    May 10, 2016 at 9:54
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    $\begingroup$ 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? $\endgroup$ May 10, 2016 at 10:53
  • $\begingroup$ @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... $\endgroup$
    – DonAntonio
    May 10, 2016 at 11:02

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