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.
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.
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.
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.
If nothing else is desired, then you could give the answer $ S_1 \times S_1 \neq S_2 $ so the statement is false.
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.