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

  • $\begingroup$ You do know that $A_n$ is usually a simple group? $\endgroup$ – Angina Seng May 13 '17 at 6:21
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    $\begingroup$ You've got to take care and not confuse "for some n" with "for any n". $\endgroup$ – ancientmathematician May 13 '17 at 6:27
  • $\begingroup$ @LordSharktheUnknown..yes i know that $A_n$ is simple group $\endgroup$ – user293581 May 13 '17 at 6:29
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    $\begingroup$ 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? $\endgroup$ – Kaj Hansen May 13 '17 at 6:40
  • $\begingroup$ $(d)$ is trivial, if you think about. You only need to know that $A_n$ is a normal subgroup of $S_n$ of index $2$. $\endgroup$ – 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.

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  • $\begingroup$ @daruma..thank you lot $\endgroup$ – user293581 May 14 '17 at 7:18

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