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How to prove that the only subgroup of the symmetric group $S_n$ of order $n!/2$ is $A_n$?

Why isn't there other possibility?

Thanks :)

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    $\begingroup$ 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. $\endgroup$ – Mariano Suárez-Álvarez Mar 14 '11 at 21:00
  • $\begingroup$ I am terribly sorry. But I don't know how to reedit it. $\endgroup$ – ShinyaSakai Mar 14 '11 at 21:07
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    $\begingroup$ There should be a link below the [abstract algebra] tag that says "edit". Click it, and you can edit. $\endgroup$ – Arturo Magidin Mar 14 '11 at 21:11
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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 $f=A_n$.

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  • $\begingroup$ Very detailed~ thank you very much~ $\endgroup$ – ShinyaSakai Mar 15 '11 at 21:28
  • $\begingroup$ $S_n$ in the last line, not $S_2$. Very nice solution. $\endgroup$ – ReverseFlow Sep 7 '14 at 23:41
  • $\begingroup$ @Genomeme: thanks, corrected $\endgroup$ – user8268 Sep 8 '14 at 21:14
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    $\begingroup$ @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. $\endgroup$ – jstnchng Nov 30 '14 at 18:49
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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.

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  • $\begingroup$ It is really a smart shortcut for the special case of $n \geq 5$~ Thank you very much~ $\endgroup$ – ShinyaSakai Mar 15 '11 at 21:30
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    $\begingroup$ 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 $S_n$...... $\endgroup$ – Vladhagen Oct 30 '13 at 19:42
  • $\begingroup$ @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 $G = S_n$. $\endgroup$ – Stephanie Oct 8 '19 at 13:14
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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 $2$.

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  • $\begingroup$ How do we know $(132)\notin H$? $\endgroup$ – Lotte Jun 8 '19 at 23:40
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    $\begingroup$ @Lotte If $(132)\in H$ as H is group $(132)^{-1}=(123)\in H$ But which is not . $\endgroup$ – idon'tknow Aug 22 '19 at 16:00
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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.

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  • $\begingroup$ How do 3-cycles have an odd order? Or were you referring to something else? $\endgroup$ – Tomás Palamás Oct 30 '18 at 20:14
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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.

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