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?


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.

  • $\begingroup$ 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

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 $$P(D_{12} \oplus \mathbb{Z}_2) \neq P(D_3 \oplus D_4).$$

  • 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

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.

  • $\begingroup$ $+1{}{}{}{}{}{}{}{}{}$ $\endgroup$ – 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

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