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My awesome math prof posted a practice midterm but didn't post any solutions to it :s Here is the question.

Let $G$ be a group and let $H$ be a subgroup of $G$.

  • (a) TRUE or FALSE: If $G$ is abelian, then so is $H$.
  • (b) TRUE or FALSE: If $H$ is abelian, then so is $G$.

Part (a) is clearly true but I am having a bit of difficulty proving it, after fulling the conditions of being a subgroup the commutative of $G$ should imply that $ab=ba$ somehow.

Part (b) I am fairly certain this is false and I know my tiny brain should be able to find an example somewhere but it is 4 am here :) I want to use some non-abelian group $G$ then find a generator to make a cyclic subgroup of $G$ that is abelian.

Any help would be appreciated, I have looked in my book but I can't seem to find for certain what I am looking for with what we have cover thus far.

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  • $\begingroup$ Thank you guys so much, we have done very little with groups and had no examples with really concrete answers yet. i just wanted to see some of examples of why it was true/false! $\endgroup$
    – Faust
    Commented Feb 7, 2013 at 13:08
  • $\begingroup$ In what way are these questions obvious if you cannot prove them? $\endgroup$ Commented Feb 13, 2013 at 20:23

7 Answers 7

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(a) Take elements $a,b \in H$. Then in particular, $a,b \in G$. Since $G$ is abelian, .... continue from here ;).

(b) What about $H=\{1\}$?

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For b, consider the non-ableian group of order $8$, $$Q_8=\langle i,j,k\mid i^2=j^2=k^2=ijk\rangle$$ and $H=\{\pm1,\pm i\}$ is a sugroup of it which is abelian.

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  • $\begingroup$ Very nice example! $\endgroup$
    – amWhy
    Commented Feb 7, 2013 at 16:02
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Concerning (a), the first times you encounter it, an a fortiori argument can be tricky, despite its simplicity. You know $a b = ba$ holds for all elements $a, b \in G$, and the elements of $H$ are just some of the elements of $G$.

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    $\begingroup$ If all my friends are nice people, those among them which like football are surely nice too! What is tricky is knowing what is meant by a fortiori argument :-) $\endgroup$ Commented Feb 13, 2013 at 20:26
  • $\begingroup$ (And trickiest of all is that the conclusion is true even if I know that none of my friends like football!) $\endgroup$ Commented Feb 13, 2013 at 20:27
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(b) Take the group of $(n \times n)$-matrices with $\mathbb{R}$-coefficients with usual matrix multiplication as G and let H be the subgroup of diagonal matrices. H ist abelian, but G is not abelian.

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(a) is easy and true since $G$ is abelian and $H$ is a subset of $G$. (b) take the symmetric group on 3 elements. It is nonabelian and has 1

abelian subgroup the alternating group of order 3 and 3 abelian subgroups of order 2.

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For things like (b) you could remember some examples of groups that are always abelian, for example groups of order 2 or 4 are always abelian, so if for some known groups you must know that are non abelian finite (symmetry groups or so...) you can find subgroups of this order you will have your counterexamples.

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For b) let $H$ be Abelian and $K$ be a nonabelian group. $G=H \times K$

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