I don't really understand factor groups in group theory (abstract algebra). I was given two things to think about:

(1) Is every factor group of an abelian group abelian? Why or why not?

(2) Is every factor group of a nonabelian group nonabelian? Why or why not?

I'm thinking no to the second one. If a nonabelian group can have an abelian subgroup, etc. But I'm not sure. Is there something I'm missing about factor groups?

Any help/hints would be most welcoming. ^_^

  • $\begingroup$ Not addressing your question directly, but perhaps a point useful to you in the future: by this year, "quotient group" is more widely understood than "factor group", which by now is somewhat anachronistic, although at some point decades ago it was very common. One reason to prefer "quotient" is that it may better suggest what's going on, and comparisons to other situations. "Factor" is quite a bit more ambiguous, even though when I was young I was accustomed to hearing it. $\endgroup$ – paul garrett Sep 15 '13 at 22:14
  • $\begingroup$ That is useful. It clarifies a few things. Thanks! $\endgroup$ – Desperate Fluffy Sep 15 '13 at 22:57

HINT: If $G=H\times K$, then $H$ is isomorphic to $H\times\{1_K\}$, which is a subgroup of $G$.

  1. What does this tell you about $H$ if $G$ is Abelian?
  2. What happens if $H$ is Abelian and $K$ isn’t? Remember, $K$ is also isomorphic to a subgroup of $G$.
  • $\begingroup$ Sorry if this is a stupid question, but what is $\{1_K\}$? $\endgroup$ – Desperate Fluffy Sep 15 '13 at 22:57
  • $\begingroup$ @DesperateFluffy: $1_K$ is the identity element of the group $K$; $\{1_k\}$ is the subset of $K$ whose only element is the identity, i.e., the trivial subgroup of $K$. $\endgroup$ – Brian M. Scott Sep 15 '13 at 22:58
  • $\begingroup$ ...that makes sense, thank you. $\endgroup$ – Desperate Fluffy Sep 15 '13 at 22:59
  • $\begingroup$ Then, if $G$ is abelian, $H$ is abelian. But if $H$ is abelian and $K$ isn't, $G$ wouldn't be abelian? $\endgroup$ – Desperate Fluffy Sep 15 '13 at 23:00
  • $\begingroup$ @DesperateFluffy: That’s right: in that case $G$ has a non-Abelian subgroup $K'$ isomorphic to $K$, so there are $x,y\in K'\subseteq G$ such that $xy\ne yx$, and that clearly implies that $G$ isn’t Abelian. $\endgroup$ – Brian M. Scott Sep 15 '13 at 23:02

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