# A question about symmetric and alternating group

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 $$n\geq1$$

• You do know that $A_n$ is usually a simple group? – Angina Seng May 13 '17 at 6:21
• You've got to take care and not confuse "for some n" with "for any n". – ancientmathematician May 13 '17 at 6:27
• @LordSharktheUnknown..yes i know that $A_n$ is simple group – user293581 May 13 '17 at 6:29
• We can only form quotient groups using normal subgroups. I.e. valid quotients of $S_n$ are of the form $S_n/N$ for a normal $N \leq S_n$. What are the normal subgroups for $S_n$? The associated quotient groups? – Kaj Hansen May 13 '17 at 6:40
• $(d)$ is trivial, if you think about. You only need to know that $A_n$ is a normal subgroup of $S_n$ of index $2$. – Dietrich Burde May 13 '17 at 8:42

(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.