# There is no simple group of order $144$

There is no simple group of order $$144$$

I have a question to the proof of the statement above (from the book J. Gallian, Contemporary abstract algebra), it is about the index theorem, so I give first the theorem

The Index Theorem:

If $$G$$ is a finite group and $$H$$ is a proper subgroup of $$G$$ such that $$|G|$$ does not divide $$|G:H|!$$, then $$H$$ contains a nontrivial normal subgroup of G. In particular, $$G$$ is not simple.

$$\bullet 3$$rd line: How does the index theorem eliminate the case $$n_3=4$$ ? • $\;n_3=4\implies$ there exists a subgroup of index $\;4\;$ ,which cannot be since $\;|G|=144\nmid 4!\;$ . The same thing at the end of the proof. – Timbuc Jan 5 '15 at 14:16
• @Timbuc sorry I have to delete the second question, because it is clear now, but why $n_3\implies$ there exists a subgroup of index $4$ – inequal Jan 5 '15 at 14:21

If $n_3=4$, then a Sylow 3-subgroup $H$ would have a normalizer $N(H)$ with index 4 (the number of Sylow $p$-subgroups is equal to the index of a normalizer of a $p$-subgroup). But you cannot have $N(H)$ with index 4 since $|G|=144=2^4\cdot 3^2$ does not divide $[G:N(H)]!=4!=2^3\cdot 3$ (the index theorem would then imply that $G$ has a proper nontrivial normal subgroup contradicting the simplicity assumption).

Again if $[G:N(H \cap H')] \leq 4$, then $|G|$ does not divide $[G:N(H \cap H')]!$ ($=4!=24,3!=6,2!=2,$ or $1!=1$) so $G$ has a proper nontrivial normal subgroup (contradiction).

Addendum: Let $n_p$ be the number of Sylow $p$-subgroups. Let $H$ be a Sylow $p$-subgroup. I claim that $n_p=[G:N(H)]$ (the number of Sylow $p$-subgroups matches the index of the normalizer of a $p$-subgroup).

This follows immediately from the "orbit-stablizer" theorem. Notice that $G$ acts on the set of Sylow $p$-subgroups via conjugation: $g \cdot H = gHg^{-1}$ is a Sylow $p$-subgroup for any $g\in G$. Why? Because conjugation preserves orders so a conjugate subgroup of a Sylow $p$-subgroup is still a Sylow $p$-subgroup.

Next, all Sylow $p$-subgroups lie in a single orbit. This is commonly referred to as the second Sylow theorem (any two Sylow $p$-subgroups are conjugate). Thus the orbit of $H$ (under this action) is the set of all Sylow $p$-subgroups. So the size of the orbit of $H$ is $n_p$.

Notice that $xHx^{-1}=H$ if and only if $x \in N(H)$. Thus the stabilizer of $H$ is exactly $N(H)$.

Finally, the orbit-stabilizer theorem says that the size of an orbit times the size of a stabilizer is equal to the size of the group. Thus $|G|=n_p \cdot |N(H)|$. But Lagrange's theorem says $|G|=[G:N(H)] \cdot |N(H)|$. Thus $n_p=[G:N(H)]$.

• ''the number of Sylow p-subgroups is equal to the index of a normalizer of a p-subgroup''. That solves the problem, any proof of it ? – inequal Jan 5 '15 at 14:25
• Thanks a lot, to put in a nutshell, can I say that the map $xN(H)\to xHx^{-1}$ is well-defined and bijective ? – inequal Jan 5 '15 at 15:05
• Yes, for a Sylow $p$-subgroup, that a bijection. That would work (and would be more to the point!). – Bill Cook Jan 5 '15 at 17:17

Let $P$ denote a 3-Sylow subgroup of $G.$ If $n_3 = 4,$ then $|G|$ does not divide 4!. So by Index theorem, $P$ contains a non-trivial normal subgroup of $G.$ This will force that $P$ is normal in $G,$ contradicting the fact that $[G : N_G(P)] = 4.$

• but how can you assume that if $n_3=4$, then there's a subgroup of $G$ with $4$ elements ? – inequal Jan 5 '15 at 14:34
• @inequal: I couldn't understand your question. Where did I mentioned that there is a subgroup with 4 elements? – Krish Jan 5 '15 at 14:37
• I meant the index, not element, apologize – inequal Jan 5 '15 at 14:39
• this is a standard fact. see the proof of Sylow's Third Theorem on the same book, it is mentioned there (in the last line). – Krish Jan 5 '15 at 14:43