# $A_n$ is the only subgroup of $S_n$ of index $2$.

How to prove that the only subgroup of the symmetric group $$S_n$$ of index 2 is $$A_n$$?

Why isn't there other possibility?

Thanks :)

• Please make the body of your posts self-contained. The title is an indexing feature, and should not be an integral part of the message. Think of it as the title of a book on the spine; it's there to let people know what the post is about, not to impart information without which you cannot understand what is happening. Commented Mar 14, 2011 at 21:00
• I am terribly sorry. But I don't know how to reedit it. Commented Mar 14, 2011 at 21:07
• There should be a link below the [abstract algebra] tag that says "edit". Click it, and you can edit. Commented Mar 14, 2011 at 21:11

As mentioned by yoyo: if $H\subset S_n$ is of index 2 then it is normal and $S_n/H$ is isomorphic to $C_2=\{1,-1\}$. We thus have a surjective homomorphism $f:S_n\to C_2$ with kernel $H$. All transpositions in $S_n$ are conjugate, hence $f(t)\in C_2$ is the same element for every transposition $t\in S_n$ (this uses the fact that $C_2$ is commutative). $S_n$ is generated by transpositions, therefore $C_2$ is generated by $f(t)$ (for any transposition $t\in S_n$), therefore $f(t)=-1$, therefore ker $f=A_n$.

• @user8268 Can you explain how $f(t)\in C_2$ is the same element for every transposition $t\in S_n$, and how $S_n$ is generated by transpositions? Thank you. Commented Nov 30, 2014 at 18:49
• @jstnchng Suppose $0 = f(t) \neq f(t') = 1$ for transpositions $t, t' \in S_n$. Then $1 = f(t) + f(t') = f(tt') = f(tgtg^{-1})$ for some $g \in S_n$. But then $1 = 0 + 0$ is a contradiction. Commented Apr 12, 2021 at 22:07
• @jstnchng This is probably too late to be helpful, but if someone has the same question, a way to see that $S_n$ is generated by transpositions is to recognize that every permutation can be written as a product of disjoint cycles, and every cycle can be written as a product of transpositions. In fact, you can show you don't even need every transposition. For instance, $\{ (12),(23),\dots,(n-1\ n)\}$ is a generating set. So is $\{ (12),(13),(14),\dots,(1n)\}$. Commented Jul 6 at 4:48

Other Way :

$$A_n$$ is generated by all $$3$$-cycles in $$S_n$$.

If $$H\neq A_n$$ and $$|S_n:H|=2$$ then at least one 3-cycle is not in $$H.$$

WLOG assume say $$(123)\notin H$$ so $$H,(123)H,(132)H$$ are 3 distinct cosets which is a contradiction to the fact that $$H$$ has index $$2$$.

• How do we know $(132)\notin H$? Commented Jun 8, 2019 at 23:40
• @Lotte If $(132)\in H$ as H is group $(132)^{-1}=(123)\in H$ But which is not . Commented Aug 22, 2019 at 16:00
• How do you know these are in fact distinct sets? Commented Jan 11, 2022 at 1:59
• Coset are partition of group. So by definition they are distinct. Let me know if you have any doubts Commented Jan 11, 2022 at 5:06
• why would at least one 3-cycle not be in $H$ ? because if we assume all 3-cycles are in $H$, we can only know that $A_n \subseteq H$ right ?but I don't see why this would imply that $A_n = H$. Commented Dec 20, 2022 at 9:59
• Subgroups of index two are normal (exercise).
• $$A_n$$ is simple, $$n\geq 5$$ (exercise).

If there were another subgroup $$H$$ of index two, then $$H\cap A_n$$ would be normal in $$A_n$$, contradiction.

[To see that $$A_n\cap H\neq\{1\}$$ as brought up in a comment, note that if this were the case, then $$S_n\backslash(A_n\cup H)=\{x\}$$ for some single element $$x\in S_n$$. Both $$A_n$$ and $$H$$ are normal, so $$x\in Z(S_n)=\{1\}$$, impossible.]

The better answer is probably the accepted one: The sign homomorphism is the only non-trivial homomorphism $$\phi:S_n\to \{\pm1\}$$. This uniquely identifies $$A_n$$ as the only index 2 subgroup of $$S_n$$.

• How do you know that the intersection is not trivial? {1} is normal in every group and does not contradict simplicity. There is a way around this using conjugacy in $S_n$...... Commented Oct 30, 2013 at 19:42
• @Vladhagen you define the normal subgroup $G$ as being non-trivial. If $G \preccurlyeq S_n$ such that $[G:S_n]=2$ and $G \ne A_n$ then $[A_n \cap G:S_n]=2$, but $A_n$ is simple and obviously $A_n \cap G \preccurlyeq A_n$ so $A_n \cap G = A_n$ or identity. If identity, then $G$ contains identity and a transposition, but then $G$ isn't normal. So the intersection must be $A_n$, and by the order divisor theorem $\# A_n = n!/2 \mid \# G$ and $\# G \mid \# S_n = n!$ so $G = S_n$. Commented Oct 8, 2019 at 13:14
• Laconically written. To expand on the bracketed paragraph: if $A_n \cap H = \{e\}$, then $(A_n \cap H)^c = \{x\}$ follows simply by counting. For $x$ as given, $yxy^{-1} = z$ for some $z \in A_n \implies x = y^{-1}zy \in A_n$ by normality, which is a contradiction. Similarly if $yxy^{-1}$ were in $H$. Thus, for every $y$, $yxy^{-1} \not \in A_n \cup H \implies yxy^{-1} = x$. Therefore, $x \in Z(S_n)$. Commented Dec 8, 2023 at 18:44
• $A_n$ is simple for $n \neq 4$, but what about the $n=4$ case. Is there any way to extend this argument? Commented Jul 6 at 5:53

I realize this question is rather old, but if this comes from Hungerford's book, there is a specific way he wants us to solve this problem, so I am providing this for the benefit of those working through Hungerford. First, one must recall that subgroups of index $2$ are normal. Hungerford's suggestion is to use the following fact:

Let $r,s$ be distinct elements of $\{1,2,...,n\}$. Then $A_n$ ($n \ge 3$) is generated by the $3$-cycles $\{(rsk) ~|~ 1 \le k \le n, k \ne r,s\}$.

Keeping that in mind, we first prove the two easy facts about subgroups of index $2$

Lemma (1): If $H$ is a subgroup of index $2$ in $G$, then $H$ contains the square of every element in $G$.

Proof: Let $g \in G$ be arbitrary. Then by Lagrange's theorem, $(gH)^2 = H$ or $g^2H=H$, happening if and only if $g^2 \in H$.

Lemma (2): If $H$ is a subgroup of index $2$ in $G$, then $H$ contains all elements of odd order.

Proof: Suppose that $g \in G$ is a element of order $2k+1$ for some $k \in \mathbb{N}$. Then $H = g^{2k+1}H = gH \cdot (g^2)^k H = gH$, where $(g^2)^k H = H$ by lemma (1). This, of course, means $g \in H$.

From these two lemmas, I think it is pretty clear how one ought to use the given fact: if $H \le S_n$ were a subgroup of index $2$, then it would consist of all $3$-cycles, since they have odd order. But by closure this means that $A_n \le H$. Since they have the same index, they must also have the same order, implying that they are equal because we are dealing with finite sets.

• How do 3-cycles have an odd order? Or were you referring to something else?
– user480875
Commented Oct 30, 2018 at 20:14
• @Maximalista 3-cycles have order 3. On the other hand, a 3-cycle is an even permutation. Commented Jan 1, 2021 at 10:48

The quotient map for $$A_n$$ is a surjective homomorphism to $$C_2$$.

Any other index two subgroup $$H$$ of $$A_n$$ gives you a distinct surjective homomorphism to $$C_2$$.

Therefore taking the product of these we obtain a surjective homomorphism $$S_n$$ to $$C_2 \times C_2$$. But then this has kernel of order $$\frac{n!}{4}$$, by the first iso theorem, and thus the image of $$A_n$$ cannot be 1 or $$C_2 \times C_2$$, so must be of order 2.

But this implies that $$A_n$$ has an index two subgroup, a contradiction as $$A_n$$ is simple for $$n \geq 5$$, and when $$n = 4$$, $$A_4$$ has no order 6 subgroup.

Let $$n\geq 2$$ and $$H \leq S_n$$ be such that $$\mid S_n:H \mid=2$$. Then $$H \trianglelefteq S_n$$ and $$S_n/H$$ being isomorphic to $$\Bbb Z/2\Bbb Z$$, is cyclic. Consider the natural surjection $$\pi :S_n\to S_n/H$$. Since $$S_n/H$$ is abelian, the commutator subgroup $$[S_n,S_n]\subseteq \ker(\pi)=H$$. We know $$[S_n,S_n]=A_n$$, hence $$A_n \subseteq H$$. Clealy $$\mid A_n \mid=\mid H \mid$$, as both of them are subgroups of a finite group of equal index. Therefore we have $$H=A_n$$.

Let $$H\le S_n$$ such that $$[S_n:H]=2$$ and $$H\ne A_n$$. Denoted $$d:=|H\cap A_n|$$, we get: $$|HA_n|=\frac{(n!/2)^2}{d} \tag1$$ By Lagrange, $$d\mid n!/2$$ and, since $$HA_n\le S_n$$ (as $$A_n$$ is normal in $$S_n$$) also $$\bigl((n!/2)^2/d\bigr)\mid n!$$. Therefore: $$\frac{n!}{2}=kd \tag2$$ and $$1=l\frac{n!}{4d} \tag3$$ for some positive integers $$k$$ and $$l$$. From $$(2)$$ and $$(3)$$ follows $$kl=2$$, and hence $$k=1$$ or $$k=2$$. But since $$H\ne A_n$$, it must be $$d\ne n!/2$$ and hence, by $$(2)$$, $$k\ne 1$$. So $$k=2$$ and hence, again by $$(2)$$, $$d=n!/4$$, whence by $$(1)$$ $$|HA_n|=n!$$, and finally $$HA_n=S_n$$: contradiction, by the argument in this post. Therefore, necessarily $$H=A_n$$.