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In our group theory classes at the university, we devoted several seminars to derive the uniqueness (up to isomorphism) of a simple group of order $168$. However, i am stuck on last steps, that prove that every group of this order must be isomorphic to PSL($2,7$). Here are the facts that i have and know the proof of:

  1. $G_{168}$ doesn't have a sugbroup with index $\leq 6$
  2. $n_7=8$, $|N_{G_{168}}(Syl_7)|=21$ and there are $48$ elements of order $7$
  3. $N_{G_{168}}(Syl_7)$ is a maximal subgroup of $G_{168}$, isomorphic with non-abelian group of the same order (there is only one such group uo to isomorphism)
  4. $n_3=28$, $|N_{G_{168}}(Syl_3)|=6$ and there are $56$ elements of order $3$
  5. $|N_{G_{168}}(Syl_3^{(1)})\cap N_{G_{168}}(Syl_3^{(2)})| \in \lbrace 1,2\rbrace$ for some $2$ non-equal $3$-Sylow subgroups
  6. $N_{G_{168}}(Syl_3)\cong D_3$, where $D_n$ is dihedral group with $2n$ elements
  7. Orders of elements in $G_{168}$ are $1,2,3,4,7,8$. There are no elements of different orders
  8. There are exactly $63$ elements of order $2,4,8$
  9. $n_2=21, |N_{G_{168}}(Syl_2)|=8$
  10. Two non-equal normalizers of $7$-Sylow subgroups have intersection of order $3$. Every $3$-subgroup is an intersection of some two non-equal normalizers of $7$-Sylow subgroups

Now, the majority of these facts follows one from another and we use them to deduce the following fact. First we define projective line field $\mathbb P^1(\mathbb F_7) =\mathbb F_7\cup \lbrace +\infty \rbrace$, where $\infty$ satisfies: $1/0=\infty, 1/ \infty=0, x\cdot \infty=\infty, x+\infty=\infty$. We then say, that $\lbrace H_0,H_1,...,H_6,H_\infty \rbrace$, where all of the elements are normalizers of different $7$-Sylow subgroups, can be identified with our projective line field, when we act on this set by some element $\beta$ of order $7$ in $H_\infty$. We then deduce, that $\beta$ acts is linear map $x\mapsto x+1$. We then take $A=H_0\cap H_\infty$ 3-Sylow subgroup and generator $\alpha$ of A and element $\gamma$ of order $2$ in $N_{G_{168}}(A)\cong D_3$. We say, that $\alpha,\gamma$ act as $\alpha: x\mapsto 2x, \gamma: x\mapsto -x^{-1}$ on our projective line field. It is easy to see, that $\alpha,\beta,\gamma$ generate $G_{168}$, using third fact mentioned above. Now we say, that $\alpha,\beta,\gamma$ act the same way $ \begin{pmatrix} 1 & 1 \\ 0 & 1 \\ \end{pmatrix}, \begin{pmatrix} 4 &0 \\ 0 & 2 \\ \end{pmatrix}, \begin{pmatrix} 0& 6 \\ 1 & 0 \\ \end{pmatrix} \in PSL(2,7)$ do on our projective line and from that somehow deduce they are isomorphic. I am quite dumbfounded, because i do not see how it follows. If the action of $G_{168}$ on our $7$ normalizers was effective, meaning the kernel is trivial, it would follow easily, because $G_{168}$ then would be isomorphic to some subgroup of $S_8$ and $PSL(2,7)$ would be homeomorphic to some subgroup $S_8$, that contains the group, isomorphic to $G_{168}$. It would remain to say that the orders are both $168$, so we have isomorphisms everywhere. However, i am not sure, if it is even true, that $G_{168}$ acts effectively.

silversurficus
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1 Answers1

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Prior thanks to testaccount, because i completely forgot about simplicity.The subgroup $G_{168}$ is simple from how we defined it, so it has no non-trivial normal subgroups. However, the kernel of an action is always normal, moreover, $G_{168}$ acts on $H_i$'s non-trivially, so from the simplicity we get, that kernel is actually a trivial subgroup, meaning the action is effective.

Now, every action of a group on some set induces a homomorphism, so we know, that $\exists \varphi: G_{168}\rightarrow H \leq S_8$ - exists a surjective homomorphism between our group $G_{168}$ and a subgroup $H$ of symmetrical group $S_8$. Because the action is effective, it is also injective, so $G_{168}\cong H$ and $H$ is also generated by $\alpha,\beta,\gamma$, when we view them as permutations. By the same reasoning we deduce existence of homomorphism $\varphi':PSL(2,7)\rightarrow H'\leq S_8$. Because $PSL(2,7)$ contains elements, that act the same way $\alpha,\beta,\gamma$ do and $G_{168}$(and therefore $H$) is generated by them, we get that $H\leq H'$, however $|H'| \leq |PSL(2,7)|=168$, so we get $|H'|=|H|=|G_{168}|=168$, meaning $\varphi'$ is actually injective, so an isomorphism. From this we get the desired result

silversurficus
  • 340