How to prove $S_3 \times \Bbb{Z}_4$ and $S_4 $ are not isomorphic groups ?
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Please write it like this: $S_3 \times \mathbf{Z}_4$ and $S_4$, in both the question and the title. – Geoffrey Trang Aug 15 '19 at 21:30
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Thank you, but in title I can't change. – THIRUMAL 5688 Aug 15 '19 at 21:46
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2Another route: the only cyclic order four subgroups of $S_4$ are generated by $4$-cycles, and none of them are normal like $\Bbb Z_4$ is within $S_3\times \Bbb Z_4$. – anon Aug 16 '19 at 00:51
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Hint: There a bunch of ways to do this, but one such way is to notice that $S_3 \times \mathbb{Z}_4$ has a cyclic subgroup of order $12$ and $S_4$ does not.
Dionel Jaime
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Another hint:
$S_4$ has $9$ elements of order $2$ whereas $S_3 \times \Bbb Z_4$ has $7$ elements of order $2$
Chinnapparaj R
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Or, for people who are too lazy to count that high, $S_3\times\mathbb Z_4$ has only two elements of order $3$, whereas $S_4$ has more than two (in fact eight, but counting up to $3$ suffices to prove non-isomorphism). – Andreas Blass Aug 16 '19 at 02:00
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The symmetric group $S_4$ has trivial center, i.e., $Z(S_4)=1$, whereas $$ Z(S_3\times C_4)=Z(S_3)\times Z(C_4)\cong 1\times C_4\cong C_4. $$
Dietrich Burde
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Find an isometry-invariant property that differs between them.
Acccumulation
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@ThirumalPachaiyappan Then that proves they're not isomorphic. If there were an isometry between them, it would send the order-24 element to an order-24 element. Although I believe it's order-12m not 24. – Acccumulation Aug 15 '19 at 21:34
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1@ThirumalPachaiyappan Actually, neither group has an element of order 24, since they are non-abelian and so cannot be cyclic. Both groups have order 24. – Geoffrey Trang Aug 15 '19 at 21:38
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4"Isometry" (same metric) is not the usual term here; isomorphism is more usual. – Arturo Magidin Aug 15 '19 at 21:48