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Why $S_4$ is not isomorphic to $D_{12}$? where $S_4$ is symmetric group and $D_{12}$ is dihedral group of order $12$.here number of element are same $S_4$ has $4!=24$ elements and $D_{12}$ has also $2n$ means $12(2)=24$ elements. but i can't find that property which holds in one of this group but another doesn't has that property.please help me thanks in advance.

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If I'm not mistaken, $S_4$ has no element of order $12$ in it, whereas $D_{12}$ has. –  k.stm Jan 29 '13 at 11:07
    
how can i justify this thing? –  Trivedi Jan 29 '13 at 11:23
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3 Answers

up vote 4 down vote accepted

Can an element in $S_4$ have order 12? Can you think of an element in $D_{12}$ or order 12? What can you conclude?

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ya i got this hint from my teacher but how can i conclude that thing bcz there are 24 permutation and in $D_{12}$ there are 12 rotation and many daigonals so what is the short justification? –  Trivedi Jan 29 '13 at 11:16
    
Is there another property which holds in $S_4$ but not in $D_{12}$? –  Trivedi Jan 29 '13 at 11:18
    
Can u give me an element of $D_{12}$ which has order 12? –  Trivedi Jan 29 '13 at 11:27
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... do you know the definition of $D_n$? –  Martin Brandenburg Jan 29 '13 at 11:28
    
i like both the answers... –  Trivedi Feb 9 '13 at 15:10
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If you already know some basic stuff about the symmetric and dihedral groups (or else you can try to prove them: it's fun) , you can try the following approach:

For $\,2<n\in\Bbb N\,$ :

$$\begin{align*}(1)&\text{ If}\;\;D_n=\langle\,a\,,\,b\;;\;a^2=b^n=1\;,\;aba=b^{n-1}=b^{-1}\rangle,\;\text{then for even}\,\,n\;\;Z(D_n)=\{1\,,\,b^{n/2}\}\\{}\\(2)&Z(S_n)=\{1\}\end{align*}$$

From the above it follows at once that $\,S_4\ncong D_{12}\,$

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can u tell me which property is break here? –  Trivedi Jan 29 '13 at 12:33
    
The groups don't have isomorphic centers...what else? –  DonAntonio Jan 29 '13 at 15:04
    
i like both the answers... –  Trivedi Feb 9 '13 at 15:11
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For even $n$, the abelianization of $D_n = \langle r,s : r^n = s^2 = (rs)^2 = 1 \rangle$ is $\langle r,s : rs=sr, r^2=s^2 = 1 \rangle = \mathbb{Z}/2 \times \mathbb{Z}/2$, but the abelianization of $S_m$ is always $\mathbb{Z}/2$ for $m>1$.

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if u explain me i will try to understand this definition. –  Trivedi Jan 29 '13 at 11:37
    
Can i say that $S_4$ is not Abelian but $D_{12}$ is abelian? –  Trivedi Jan 29 '13 at 12:39
    
$D_{12}$ is not abelian. –  Martin Brandenburg Jan 29 '13 at 14:55
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