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True or false? Give reason.

$S_m\times S_n\simeq S_{m+n}$.

I know this is not true but I don't know how to prove it.

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    Interesting Exercise: Show there is a subgroup of $S_{m+n}$ isomorphic to $S_m \times S_n $ and thus deduce $m!n!|(m+n)!.$2012-04-05

4 Answers 4

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If two groups are isomorphic, then there exists a bijection between them and thus they have the same order. The order of $S_m \times S_n$ is $|S_m| \cdot |S_n| = m! \cdot n!$, and the order of $S_{m+n}$ is $(m+n)!$.

Try to show that $S_m \times S_n$ always has smaller order than $S_{m+n}$. Then since the two groups have different order, they cannot be isomorphic.

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Cardinality of $S_{m+n}$ / cardinality of $S_m \times S_n$ = $\frac{(m+n)!}{m!n!}$ = $^{m+n}C_n$, which is greater than 1 if m is greater than 1.

EDIT: Meant 'm is greater than 0',thanks Derek for pointing it out.

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    Yes, if you allow $m = 0$ or $n = 0$, then for example $S_0 \times S_1 \cong S_1$.. here $S_0$ is the set of bijections on the empty set (symmetric group on $0$ letters), which has exactly one element, the empty function. But I think a lot of times people define $S_n$ only for $n \geq 1$2012-04-05
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If nothing else is desired, then you could give the answer $ S_1 \times S_1 \neq S_2 $ so the statement is false.

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To all the answers given so far, let's add that if it were $S_m\times S_n\simeq S_{m+n}$ then $S_{m+n}$ would have lots of normal subgroups, and we know that that cannot be.