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I have a lot of difficulty understanding this proof we went over in class about classification of groups of order 12.

Let $G$ be a finite group of order $n=p^rm$ where $m \nshortmid p$.

Denote $Syl_p(G)$ as the set of all sylow p-subgroups, and $n_p(G)$ as the cardinality of $Syl_p(G)$.

(1) G has subgroups of order $p^r$. (How do we know this??)

(2) We know all sylow p-groups are conjugate and their number $n_p(G) | m$, by the 2nd and 3rd Sylow Theorems.

(3) We have that $n_p(G) \equiv 1(p). $

Then $n=12=2^2*3^1$ so $n_2(G)|3$ gives $n_2(G) \in \{1,3\}$ and $n_3(G)|4$ so $n_3(G) \in \{1,4\}$ since $2 \not\equiv 1(3)$.

Case 1: $n_3(G) = 4.$ Then the action of G by conjugation on $Syl_3(G)$ gives a homomorphism $f: G \rightarrow S_{Syl_3(G)} \simeq S_4$. $Ker(f)$ consists of $g \in G$ that normalizes all 3-Sylow subgroups. (Why?). Let $P_3$ be a 3-Sylow subgroup. Then the order of $P_3$ is 3. (I don't get that either), and $[G:N_G(P_3)] = 4.$ Thus $P_3 \subset N_G(P_3)$ gives $P_3 = N_G(P_3)$. (I'm lost here...) So $Ker(f) = \cap Syl_3(G) = \{e\}$ (the intersection of the sylow 3-subgroups is trivial. (How did we arrive here...) Thus we conclude that $G$ is isomorphic to a subgroup of $S_4$ of order 12.

To show $G$ is isomorphic to $A_4$, we have that $G$ has 8 elements of order 3: 4 3-sylow subgroups each has 3-1=2 elements of order 3. And $S_4$ has 8 3-cycles meaning f(G) contains all 3-cycles and hence the group they generate, which is $A_4$, so $A_4 \subset f(G)$. But since $|A_4| = 12 = |f(G)$ then $f(G) = A_4$ and $G \simeq A_4$. (Not understanding the first half of the proof, I do not get this part either).

Case 2: $n_3(G) = 1.$ The scope of the question is too large, I will need to post separately to understand the second half, unless the organizational rules of math.stackexchange would require that I post here. In that case, I'll edit the question.

I basically cannot follow the proof because it seems to skip too many steps; it would be helpful if anyone could elaborate case 1 with more reasoning so that I can follow it. Any help would be appreciated; I've spent hours trying to look up and decipher the proof but without any success.

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  • $\begingroup$ You don't get that the order of $P_3$ is $3$? $\endgroup$ – Derek Holt Nov 25 '15 at 23:06
  • $\begingroup$ I get that a $P_3$ would have order a power of 3, but that could be 1, 3 or 9. $\endgroup$ – user2193268 Nov 25 '15 at 23:07
  • $\begingroup$ Okay, @DerekHolt so by Lagrange, 9 cannot divide 12 and if it were 1, then we could not have 4 unique p-subgroups. $\endgroup$ – user2193268 Nov 26 '15 at 3:07
  • $\begingroup$ I think you need to understand Sylow's Theorem better before you can attempt this problem. $\endgroup$ – Derek Holt Nov 26 '15 at 8:38
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I have a lot of difficulty understanding this proof we went over in class about classification of groups of order 12.

Let $G$ be a finite group of order $n=p^rm$ where $m \nshortmid p$.

It seems that there is a mistake here one should read $p \nshortmid m$ or $gcd(p,m)=1$.

Denote $Syl_p(G)$ as the set of all sylow p-subgroups, and $n_p(G)$ as the cardinality of $Syl_p(G)$.

(1) G has subgroups of order $p^r$. (How do we know this??)

A priori, it has no reason to be true but this is a theorem known as $\textit{the first Sylow's theorem}$. It states that for any finite group $G$ and $p$ a prime number dividing $|G|$, if we write $|G|=p^rm$ with $p \nshortmid m$ then $G$ admits a $p$-Sylow (which is defined as a subgroup of $G$ whose cardinal is $p^r$).

(2) We know all sylow p-groups are conjugate and their number $n_p(G) | m$, by the 2nd and 3rd Sylow Theorems.

(3) We have that $n_p(G) \equiv 1(p). $

Then $n=12=2^2*3^1$ so $n_2(G)|3$ gives $n_2(G) \in \{1,3\}$ and $n_3(G)|4$ so $n_3(G) \in \{1,4\}$ since $2 \not\equiv 1(3)$.

Case 1: $n_3(G) = 4.$ Then the action of G by conjugation on $Syl_3(G)$ gives a homomorphism $f: G \rightarrow S_{Syl_3(G)} \simeq S_4$. $Ker(f)$ consists of $g \in G$ that normalizes all 3-Sylow subgroups. (Why?).

How is $f$ defined ? Since it is given by the conjugation action $f(g):=S\mapsto gSg^{-1}$ so that if $f(g)$ is trivial (the indentity function here), it follows that $f(g)$ sends any $S$ to $S$ since $f(g)$ also sends any $S$ to $gSg^{-1}$ this boils down to for all $S$ we have $S=gSg^{-1}$.

Let $P_3$ be a 3-Sylow subgroup. Then the order of $P_3$ is 3. (I don't get that either)

This is the very definition of a $3$-Sylow in $G$ that gives you this since $|G|=12=4\times 3$.

, and $[G:N_G(P_3)] = 4.$ Thus $P_3 \subset N_G(P_3)$ gives $P_3 = N_G(P_3)$. (I'm lost here...)

Since $G$ acts transitively by conjugation (this is your point 2 : $p$-Sylow are all conjugate to each other) we have by the first class formula :

$$\frac{|G|}{n_3(G)}=|N_G(P_3)|\text{ whence } |N_G(P_3)|=3$$

By definition $H\subseteq N_G(H)$ so that $P_3\subseteq N_G(P_3)$ since both are of cardinal $3$ they are equal.

So $Ker(f) = \cap Syl_3(G) = \{e\}$ (the intersection of the sylow 3-subgroups is trivial. (How did we arrive here...)

We already saw that $Ker(f)$ is the intersection of normalizers of $3$-Sylows. Now we just saw that the normalizer of a $3$-Sylow is exactly the $3$-Sylow itself. Now I claim that if $S$ and $T$ are two different $3$-Sylows they cannot but intersect trivially. Indeed by Lagrange $S\cap T$ has cardinal dividing $|S|=3$. So the cardinal is either $3$ or $1$, if $|S\cap T|=3$ then $S\cap T\subseteq S$ and $|S|=3$ implies that $S\cap T=S$ and then that $S\subseteq T$ which implies (with the cardinal) that $S=T$ which is impossible. It follows that $|S\cap T|=1$ i.e. $S\cap T$ is trivial.

Since $Ker(f)$ is the intersection of $4$ different $3$-Sylows it is necessarily trivial.

Thus we conclude that $G$ is isomorphic to a subgroup of $S_4$ of order 12.

To show $G$ is isomorphic to $A_4$, we have that $G$ has 8 elements of order 3: 4 3-sylow subgroups each has 3-1=2 elements of order 3. And $S_4$ has 8 3-cycles meaning f(G) contains all 3-cycles and hence the group they generate, which is $A_4$, so $A_4 \subset f(G)$. But since $|A_4| = 12 = |f(G)$ then $f(G) = A_4$ and $G \simeq A_4$. (Not understanding the first half of the proof, I do not get this part either).

Since $Ker(f)$ has been proven to be trivial it follows that $G$ is isomorphic to $f(G)$ which is a subgroup of $S_4$. Now you would like to identify the subgroup $f(G)$ you know that it contains $4$ $3$-Sylows. A $3$-Sylow is of order $3$, it contains two elements of order $3$ and one of order $1$. Since the intersection of two different $3$-Sylows is trivial and since you have $4$ $3$-Sylows you finally get in $f(G)$ exactly $8$ elements of order $3$ in $f(G)$. Since they are elements of order $3$ in $S_4$ they cannot but be $3$-cycles and since there are $8$ of them in $S_4$ they are all contained in $f(G)$. The fact that they generate $A_4$ is classical so that $A_4\leq f(G)$ and by cardinality argument they are equal.

This part is too complicated, one way to define $A_n$ is to say that this is the unique subgroup of index $2$ in $S_4$. Since $S_4$ is of order $24$ and $f(G)$ of order $12$ it follows that $f(G)$ is a subgroup of index $2$ in $S_4$ and then $A_4=f(G)$ by unicity.

Case 2: $n_3(G) = 1.$ The scope of the question is too large, I will need to post separately to understand the second half, unless the organizational rules of math.stackexchange would require that I post here. In that case, I'll edit the question.

I basically cannot follow the proof because it seems to skip too many steps; it would be helpful if anyone could elaborate case 1 with more reasoning so that I can follow it. Any help would be appreciated; I've spent hours trying to look up and decipher the proof but without any success.

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  • $\begingroup$ I think it'd be clearer if your input was in plain text and the original text highlighted. $\endgroup$ – lhf Nov 27 '15 at 11:08
  • $\begingroup$ @ClémentGuérin Thanks a lot for your help! I went through all of it, slowly and I understand it now. There is one last thought left: Since they are elements of order 3 in S4 they cannot but be 3-cycles. Is it true that 4-cycle can't be of order 3? $\endgroup$ – user2193268 Dec 1 '15 at 20:51
  • $\begingroup$ @user2193268, in general, in $S_n$ a $k$-cycle is of order $k$,it is a general property that directly comes from the definition of a $k$-cycle. $\endgroup$ – Clément Guérin Dec 2 '15 at 7:31

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