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Are there isomorphic groups between $\Bbb{A}_4\times \Bbb{Z}_3,\Bbb{D}_{18},\Bbb{D}_{9}\times \Bbb{Z}_2,\Bbb{S}_3\times \Bbb{S}_3$? Where $\Bbb{D}$ is the dihedral group,$\Bbb{A}$ alternating group and $\Bbb{S}$ is the symmetric group.

I've tried the things below but I feel like I'm doing things wrongly and I don't know what to do with $\Bbb{A}_4\times \Bbb{Z}_3$.

For $\Bbb{D}_{18}$ and $\Bbb{D}_{9}\times \Bbb{Z}_2$ I've tried constructing $f$ such that $f(\sigma\rho^k) = (\sigma\rho^{(k-k\bmod 2)/2},k\bmod 2)$ and $f(\rho^{2k})=(\rho^k,0)$ and $f(\rho^{9})=(1,1)$ and $f(\rho)=(\rho,1),f(\rho^3)=(\rho^2,1),f(\rho^5)=(\rho^3,1),f(\rho^7)=(\rho^4,1),f(\rho^{11})=(\rho^5,1),f(\rho^{13})=(\rho^6,1),f(\rho^{15})=(\rho^7,1),f(\rho^{17})=(\rho^8,1)$ which should be an isomorphism but I couldn't prove it.

For $\Bbb{D}_{18}$ and $\Bbb{S}_3\times \Bbb{S}_3$, $\Bbb{D}_{18}$ has a element of order $18$ while $\Bbb{S}_3\times \Bbb{S}_3$ has at most an element of order $9$. Similar with $\Bbb{D}_{9}\times \Bbb{Z}_2$.

kingW3
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  • @Servaes My guess is that it should be group of order $2n$ otherwise the groups wouldn't have the same number of elements. – kingW3 Jan 09 '19 at 16:36

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Your idea for an isomorphism $\Bbb{D}_{18}\cong\Bbb{D}_9\times\Bbb{Z}_2$ is good. What part of proving this is an isomorphism is a problem? Try writing out what $f$ does to an element $\sigma^a\rho^b\in\Bbb{D}_{18}$ in general, in stead of defining it for each element separately. That will surely make it more clear why this should be an isomorphism.

Your last consideration is a good one; the groups $\Bbb{D}_{18}$ and $\Bbb{S}_3\times\Bbb{S}_3$ are indeed not isomorphic because $\Bbb{D}_{18}$ has an element of order $18$ while $\Bbb{S}_3\times\Bbb{S}_3$ does not. Similar arguments show that no two of these groups are isomorphic, except $\Bbb{D}_{18}$ and $\Bbb{D}_9\times\Bbb{Z}_2$. As this seems like a homework exercise, I'll leave the rest for you to figure out (for now).

Servaes
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    What about $D_{18}$ and $D_9 \times Z_2$? – Derek Holt Jan 09 '19 at 16:36
  • @DerekHolt Oops, I always get confused with dihedral groups, let me edit my answer. – Servaes Jan 09 '19 at 16:41
  • The order of $A_4$ elements seems to be at most of order $3$ and similar with $Z_3$ and $S_3$, so that the order of the element with biggest order in both $S_3\times S_3$ and $A_4\times Z_3$ can be the same $9$ so I'm not sure what to do. Also this isn't homework I'm just preparing for an exam. – kingW3 Jan 09 '19 at 16:56
  • This is not immediately relevant to the question, but neither has elements of order $9$. Given elements $a\in A$ and $b\in B$ of finite order, what is the order of $(a,b)\in A\times B$? – Servaes Jan 09 '19 at 17:03
  • More relevant to the question; how many elements of order $3$ do these two groups have? – Servaes Jan 09 '19 at 17:03
  • @Servaes It is $\operatorname{lcm}(\operatorname{ord}(a),\operatorname{ord}(b))$. So the elements $(a,b)$ are of order $3$ if $a$ is of order $1$ and $b$ is of order $3$ or if $b$ is of order 1 and $a$ is of order $3$ or $a,b$ are both of order $3$.The last question is the converse true if both groups have the same number of elements of same order for every order does it mean that they are isomorphic? – kingW3 Jan 09 '19 at 17:23
  • Great! To your last question; no, see this question. – Servaes Jan 09 '19 at 17:46
  • @kingW3 Neither $S_3 \times S_3$ not $A_4 \times Z_3$ has elements of order $9$. The element orders in both of these groups are $1,2,3,6$. – Derek Holt Jan 10 '19 at 10:27