Let $S_n$ denote the symmetric group on $n$ symbols (i.e., the group of permutations of $\left\{1,2,\ldots,n\right\}$, and $A_n$ be the subgroup of even permutations. Which of the following is true?
(a) There exists a finite group which is not a subgroup of $S_n$ for any $n \geq 1$.
(b) Every finite group is a subgroup of $A_n$ for some $n\geq1$.
(c) Every finite group is quotient group of $A_n$ for some $n \geq 1$.
(d) No finite abelian group is a quotient of $S_n$ for $n>3 $.
my attempt with my knowledge: every finite group is (isomorphic to) a subgroup of $S_n$ and $S_n$ is ismorphic subgroup of $A_{n+2}$
so Every finite group is a subgroup of $A_n$ for any $n\geq1$
but i am not sure about my answer and and other options any help me please
Best Answer
(a) As you said in your post, this is not true by Cayley's Theorem which states that every finite group is isomorphic to a subgroup of $S_n$. You can look at the left regular action of $G$ on $G$ and identify it with a subgroup of $S_n$ where $n$ is the order of the group $G$. Usually, you can embed your finite group $G$ for a smaller value of $n$ but this suffices.
(b) True. If $G$ can be embedded into $S_n$, you can embed it into $A_{2n}$. For example, if you have $C_2\cong\{e,(12)\}$ then you can embed it into $A_4$ as $\{e,(1 2)(34)\}$.
(c) As mentioned in the comments by @Lord Shark the Unknown, since $A_n$ is simple for $n>4$, this is false. You can only get the quotient groups that you get from $A_2$, $A_3$ and $A_4$.
(d) Unless, I am interpreting the question wrong, @Dietrich Burde has already given us a counterexample to this. $S_n/A_n\cong C_2$ and $C_2$ is certainly an abelian group.