The order of permutation groups and alternating groups The question was: 


  True or False: $\forall{n}\in{\mathbb{N}}$ the group $S_n$ and $A_n$ have different sizes.


My answer is False. That is since  both $A_1 =(\text{id})$ and $S_1 =(\text{id})$. 

Can any one confirm my answer please? Thank you very much.
Of course $S_n$ in general is much larger than $A_n$. $A_n$  is the group of permutations with even parity only, while $S_n$ is the group of all permutations of $\{1,\cdots,n\}$. Not all permutations are of even parity, e.g., consider
$$(1,2)\circ(1,3)\circ(1,5)$$
the product of three transpositions that give rise to a cycle. In fact any cycle of odd length does not belong to $A_n$.
Of course $S_n$ in general is much larger than $A_n$. $A_n$  is the group of permutations with even parity only, while $S_n$ is the group of all permutations of $\{1,\cdots,n\}$. Not all permutations are of even parity, e.g., consider
$$(1,2)\circ(1,3)\circ(1,5)$$
the product of three transpositions that give rise to a cycle. In fact any cycle of odd length does not belong to $A_n$.