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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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Hint: what's the cardinality of the respective sets? – InvisiblePanda Apr 5 '12 at 9:11
@rand This is where i stuck. I know the cardinality of S_m+n is (m+n) factorial. But I have no idea how to find the cardinality of S_m x S_n. – faisal Apr 5 '12 at 9:17
@faisal - The cardinality of $S_m\times S_n$ is the cardinality of the set $S_m\times S_n$ - the set of all ordered pairs $(s,t)$ such that $s\in S_m$; $t\in S_n$. – Donkey_2009 Apr 5 '12 at 9:18
@Donkey - Ok. If I get it right then m! X n! is the cardinality of S_m x S_n. – faisal Apr 5 '12 at 9:23
Interesting Exercise: Show there is a subgroup of $S_{m+n}$ isomorphic to $S_m \times S_n $ and thus deduce $m!n!|(m+n)!.$ – Ragib Zaman Apr 5 '12 at 10:24

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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It's greater than one even if $m=1$. (I guess we are asumming $m,n>0$.) – Derek Holt Apr 5 '12 at 10:44
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 $n \geq 1$ – Mikko Korhonen Apr 5 '12 at 11:45

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.

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