Goal: Show that $D_{33}$ is not isomorphic to $D_{11} \oplus Z_{3}$.
They are both non-cyclic groups of order $66$. The same orders are possible for their elements. Comparing massive Cayley Tables is impractical. Any thoughts?
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Chain Markov
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JustSomeGuy716
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2What's the centre of $D_{33}$? – Angina Seng Mar 25 '19 at 16:49
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3What about counting the elements with order $2$? – Jack D'Aurizio Mar 25 '19 at 16:51
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2Possible duplicate of Prove that $D_3 \oplus D_4$ is not isomorphic to $D_{12}\oplus\mathbb Z_2$. – Dietrich Burde Mar 25 '19 at 19:54
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2Possible duplicate of Show that $D_{11} \oplus \mathbb{Z}3 \not \cong D_3 \oplus \mathbb{Z}{11}$. – Dietrich Burde Mar 25 '19 at 19:59
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$D_{33} = C_{33} \rtimes C_2 = \langle a, b| a^{33} = b^{2} = bab^{-1}a = 0 \rangle$. One can see, that every element of it can be written as $a^{n}$ or $a^nb$. Now, lets describe all elements that commute with $b$. $ba^{n}b^{-1} = a^{-n}$ and $ba^{n}bb^{-1} = a^{-n}b$. Thus, only $e$ and $b$ commute with $b$ and so they are the only possible candidates for being in the center. But $b$ does not commute with $a$ and so is no in the center either, which makes $e$ the only element of the center. So $D_{33}$ is centerless.
Now $D_{11} \times C_3$ has $C_3$ as a non-trivial abelian direct factors. And it is known, that abelian direct factors always lie in center. So $Z(D_{11} \times C_3)$ has a non-trivial subgroup and thus is non-trivial. And so, $D_{11} \times C_3$ is not-centerless.
Thus we can conclude, that they are non-isomorphic.
Chain Markov
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@J.W.Tanner, yes, it should have definitely been $D_{11}$. $D_{33}$ in the second paragraph was at the result of a typo. I have corrected it now. – Chain Markov Mar 25 '19 at 17:44
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In general, $D_{2n+1}$ has trivial center, see here:
In particular, for $n=16$ we see that $Z(D_{33})=1$. On the other hand, $$ Z(D_{11}\times C_3)=Z(D_{11})\times Z(C_3)\cong C_3. $$
Dietrich Burde
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