5
$\begingroup$

This is a problem from the textbook, doing this for practice and not assignment.

Prove that $D_3 \oplus D_4$ is not isomorphic to $D_{12}\oplus\mathbb Z_2$.

So we know

$|D_3| = 6$ and $|D_4| = 8$ then $|D_3\oplus D_4| = 6\cdot 8 = 48,$ and similarly $|D_{12}\oplus\mathbb Z_2| = 24\cdot 2 = 48.$

So the groups are of same size. We can also see that none of the groups are cyclic since orders of the individual groups are not relatively prime.

Would there be a intuitive way of approaching this question, rather then looking at the number of elements of each order in the two groups?

$\endgroup$
8
$\begingroup$

One of the perhaps many ways to arrive at a solution is to count the elements in the center of each group. Clearly $(a,b)\in Z(G⊕H)$ if and only if $a\in Z(G)$ and $b\in Z(H)$.

$Z(D_{12})= \{a^6,e\}$, $Z(D_4)=\{a^2,e\}$ and $Z(D_3)=\{e\}$.

Now just counting elements we can see that $|Z(D_3⊕D_4)|=2$ and $|Z(D_{12}⊕\mathbb Z_2)|=4$, so clearly the groups are not isomorphic.

Here is a proof of the orders of the center of the dihedral groups for reference.

$\endgroup$
  • $\begingroup$ Thanks, that's a good solution and what I was looking for. If the two groups are isomorphic there orders of center should be the same, didn't think about that one. But shouldn't $Z(D_{12})$ = {$a^6$, e}? $\endgroup$ – user77404 Aug 8 '13 at 23:44
  • $\begingroup$ @user77404 Yes it's $a^6$ not $a^3$. Stupid dihedral groups and their inconsistent conventions! $\endgroup$ – PVAL-inactive Aug 8 '13 at 23:49
5
$\begingroup$

For any group $G$, let $P(G)$ be the probability that two elements of $G$ commute. Then $$P(D_{12} \oplus \mathbb{Z}_2)= P(D_{12}) \ \text{and} \ P(D_3\oplus D_4)= P(D_3) \cdot P(D_4).$$

With some calculations, we find $P(D_3)=\frac{1}{2}$, $P(D_4)= \frac{5}{8}$ and $P(D_{12})= \frac{3}{8}$, hence $$P(D_{12} \oplus \mathbb{Z}_2) \neq P(D_3 \oplus D_4).$$

$\endgroup$
  • 2
    $\begingroup$ +1 very interesting. I have thought that the statistics and probability are useless just in Gambling. Very nice. $\endgroup$ – mrs Aug 10 '13 at 16:32
2
$\begingroup$

Also you may consider the solutions of equation $x^2=id$ in two groups. In the first group we have $24$ elements while in the other group we have $28$. I used GAP for this aim.

$\endgroup$
  • $\begingroup$ $+1{}{}{}{}{}{}{}{}{}$ $\endgroup$ – Namaste Aug 10 '13 at 13:18
  • $\begingroup$ @amWhy: Don't worry. Just to take enough rest to be fresh here again. Back to the site so Energetic, Amy. :) $\endgroup$ – mrs Aug 10 '13 at 17:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.