$G$ is non abelian simple group of order $<100$ then $G\cong A_5$

Question is to Prove that :

Hint is to "Eliminate all orders but $60$". Which i think is not so easy to check.

First of all, I eliminate all Primes (cyclic groups) $2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97.$ (including 1)

Only numbers i am left with are, $\{4,6,8,9,10,12,14,15,16,18,20,21,22,24,25,26,27,28,30,32,33,34,35,36,38,39,40,42,44,45,46,48,49,50,51,52,54,55,56,57,58,60,62,63,64,65,66,68,69,70,72,74,75,76,77,78,80,81,82,84,85,86,87,88,90,91,92,93,94,95,96,98,99\}$

Now, I eliminate all prime squares $p^2$ (which are abelian) $4,9,16,25,36,49,64,81$

Only numbers i am left with are, $\{6,8,10,12,14,15,18,20,21,22,24,26,27,28,30,32,33,34,35,38,39,40,42,44,45,46,48,50,51,52,54,55,56,57,58,60,62,63,65,66,68,69,70,72,74,75,76,77,78,80,82,84,85,86,87,88,90,91,92,93,94,95,96,98,99\}$

Now i eliminate all prime powers $p^n$ (which have non trivial center hence not simple) $2^3=8,2^5=32$ and $3^3=27$

Only numbers i am left with are, $\{6,10,12,14,15,18,20,21,22,24,26,28,30,33,34,35,38,39,40,42,44,45,46,48,50,51,52,54,55,56,57,58,60,62,63,65,66,68,69,70,72,74,75,76,77,78,80,82,84,85,86,87,88,90,91,92,93,94,95,96,98,99\}$

Now i eliminate all groups of order $pq$ where $p$ and $q$ are distinct (they have either normal sylow p or sylow q subgroup)

From 2 - $\{2p=6,10,14,22,26,34,38, 46,58,62,74,82,86,94\}$

Only numbers i am left with are, $\{12,15,18,20,21,24,28,30,33,35,39,40,42,44,45,48,50,51,52,54,55,56,57,60,63,65,66,68,69,70,72,75,76,77,78,80,84,85,87,88,90,91,92,93,95,96,98,99\}$

From 3 - $\{ 3p=15,21,33,39,51,57,69,87,93\}$

Only numbers i am left with are, $\{12,18,20,24,28,30,35,40,42,44,45,48,50,52,54,55,56,60,63,65,66,68,70,72,75,76,77,78,80,84,85,88,90,91,92,95,96,98,99\}$

From 5- $\{5p= 10,15,35,55,65,85,95\}$

Only numbers i am left with are, $\{12,18,20,24,28,30,40,42,44,45,48,50,52,54,56,60,63,66,68,70,72,75,76,77,78,80,84,88,90,91,92,96,98,99\}$

From 7 - $\{7p= 14,21,35,49,77,91,\}$

Only numbers i am left with are, $\{12,18,20,24,28,30,40,42,44,45,48,50,52,54,56,60,63,66,68,70,72,75,76,78,80,84,88,90,92,96,98,99\}$

Remaining products $pq$ repeats.

Now i eliminate all groups of order $p^2q$ (which are not simple as they have either normal sylow p or sylow q subgroup)

From 2 - $4p=\{ 8,12,20,28,44,52,68,76,92,\}$

Only numbers i am left with are, $\{18,24,30,40,42,45,48,50,54,56,60,63,66,70,72,75,78,80,84,88,90,96,98,99\}$

From 3 - $9p= \{ 18,27,45,63,99\}$

Only numbers i am left with are, $\{24,30,40,42,48,50,54,56,60,66,70,72,75,78,80,84,88,90,96,98,\}$

From 5 - $25p =\{50,75 \}$

Only numbers i am left with are, $\{24,30,40,42,48,54,56,60,66,70,72,78,80,84,88,90,96,98,\}$

From 7 - $49p= \{ 98\}$ Only numbers i am left with are, $\{24,30,40,42,48,54,56,60,66,70,72,78,80,84,88,90,96,\}$

Now i eliminate all groups of order $pqr$, p,q,r are distinct primes (which are not simple as they have either normal sylow p or sylow q subgroup or sylow r subgroup)

Only numbers we left with are $\{24,30,40,42,48,54,56,60,66,70,72,78,80,84,88,90,96\}$

$30=2.3.5$ So, $30$ Is eliminated

$42=2.3.7$ So, $42$ is eliminated

$66=2.3.11$ So, $66$ is eliminated

$70=2.5.7$ So, $70$ is eliminated

$78=2.3.13$ So, $78$ is eliminated

Only numbers we are left with are $\{ 24,40,48,54,56,60,72,80,84,88,90,96\}$

$\textbf{EDIT}$ Assuming $G$ is simple, $|G|=24$. as $24=2^3.3$ No. of sylow 3 subgroups $1+3k$ divides $8$.thus, no of sylow $3$ subgroups has to be $4$. Suppose $n_2=3$ each sylow 3 subgroup has 2 non identity elements totally 8 non identity elements. each sylow 2 subgrousp has 7 non identity elements as we can not assume sylow 2 subgroups intersection is non trivial,$\textbf{INCOMPLETE}$

Only numbers we are left with are $\{40,48,54,56,60,72,80,84,88,90,96\}$

$40=2^3.5$ no.of sylow 5 dubgroups $1+5k$ divides $8$ Thus,sylow 5 sbgroup is unique and hence group is not simple.

Only numbers we are left with are $\{48,54,56,60,72,80,84,88,90,96\}$

$48=2^4.3$ No. of sylow 3 subgroups $1+3k$ divides $16$ No. of sylow 2 subgroups $1+2k$ divides 3 suppose $n_2=3$ and $n_3=16$ Contribution from $P_3$(sylow 3 subgroup) is 45 and contribution from $P_2$ (sylow 2 subgroup) is 3 which adds up to 48 and with identity element we have 49 elements. Thus at least one of $n_3$ or $n_2$ is $1$ So, G is not simple.

Only numbers we are left with are $\{54,56,60,72,80,84,88,90,96\}$

$54=2.3^3$ No. of sylow 3 sbgroups, $1+3k$ divides $2$ Thus sylow 3 subgroup is normal and hence group is not simple.

Only numbers we are left with are $\{56,60,72,80,84,88,90,96\}$

$\textbf{EDIT}$ : Consider group $G$ of order $56$, for this we have $56=2^3.7$. Assuming this being simple group we would end up with the case that $n_2=7,n_7=8$. $n_p$ denotes no. of sylow p subgroups. each sylow $7$ subgroup has $6$ non identity elements, totally there are $8\times 6=48$ non identity elements in all sylow $7$ subgroups.\ each sylow $2$ subgroup has $7$ non identity elements. As there is a possibility that intersection of two sylow $2$ subgroups to be non trivial, there would be one more (non identity) element different from these seven non identity elements, adding upto 8 non identity elements, with $1$ identity element and adding upto $1+8+48=57$ ($48$ elements from sylow $7$ subgroups) contradicting the cardinality of order of group $|G|=56$. Thus, either $n_2=1$ or $n_7=1$.Thus, there exists a unique sylow $2$ subgroup or a unique sylow $7$ subgroup.Thus,$G$ is not simple.

Only numbers we are left with are $\{60,72,80,84,88,90,96\}$

$80=2^4.5$ Possibilities for $n_2$ are $1,5$ possibilities for $n_5$ are $1,16$. Suppose Suppose $n_2=5$ and $n_5=16$, then $P_5$ contributes $64$ elements and atleast $16$ non identity elements from $P_2$ adding up to $80$ excluding identity. Thus G is not simple

Only numbers we are left with are $\{60,72,84,88,90,96\}$

$84=2^2.3.7$ with out much difficulty, one can see $n_7=1$ and thus, G is not simple.

Only numbers we are left with are $\{60,72,88,90,96\}$

$88=2^3.11$ with out much difficulty, one can see that $n_{11}=1$ thus G is simple.

Only numbers we are left with are $\{60,72,90,96\}$

I somehow managed to show groups of order $72,90,96$ are not simple. (My hands are paining I can not write more than this :D)

So, I am left with group of order $60$ and we have $A_5$ with $|A_5|=60$ and $G\cong A_5$

I would be thankful if someone can check whether this is clear (or) not (even with any typos) and If possible give a hint for a simple way to arrive at required result.

Thank You.

P.S : To be frank, I have no idea how to solve this before writing this. I thought i would say at-least i know cyclic groups (prime order) are not simple and leave the rest to the other users and then I realized $p^2$ are not simple and so on tried eliminating one by one. at the time i came to the case of $72,90,96$ I got fed up ad blindly decided to assume they are simple (:P). I would write about that cases in a while in detail.

P.S $2$ : Could any one please help me in concluding a group of order 24 being simple. I have edited a blunder in my argument. But could not able to proceed further.I am hoping for a proof which use counting argument and no other results :)

Solution 1:

By using some more powerful results, it is possible to do this a lot easier.

The two main ingredients for this will be the following:

Burnside's $pq$-Theorem: If only two distinct primes divide the order of $G$, then $G$ is solvable.

Burnside's Transfer Theorem: If $P$ is a $p$-Sylow subgroup of $G$ and $P\leq Z(N_G(P))$ then $G$ has a normal $p$-complement.

So when looking for a non-abelian simple group, the first result immediately tells us that we can assume at least $3$ distinct primes divide the order of $G$.

The way we will use the second result is the following:

Let $P$ be a $p$-Sylow subgroup where $p$ is the smallest prime divisor of $|G|$. We will show that if $|P| = p$ then $P\leq Z(N_G(P))$ (and then the above result says that $G$ is not simple).

To see this, we note that $N_G(P)/C_G(P)$ is isomorphic to a subgroup of $\rm{Aut}(P)$ (this is known as the N/C-Theorem and is a nice exercise), which has order $p-1$, and since the order of $N_G(P)/C_G(P)$ divides $|G|$, this means that $N_G(P) = C_G(P)$ and hence the claim (since we had picked $p$ to be the smallest prime divisor).

The same argument actually shows that if $p$ is the smallest prime divisor of $|G|$, then either $p^3$ divides $|G|$ or $p = 2$ and $12$ divides $|G|$. The reason for this is that if the $p$-Sylow is cyclic of order $p^2$ the precise same argument carries through, since in that case all prime divisors of $|\rm{Aut}(P)|$ are $p$ or smaller.

If $P$ is not cyclic then the order of $\rm{Aut}(P)$ will be $(p^2-1)(p^2 - p) = (p+1)p(p-1)^2$ and the only case where this can have a prime divisor greater than $p$ is when $p = 2$ in which case that prime divisor is $3$, and we get the claim.

In summary we get that the order of $G$ must be divisible by at least $3$ distinct primes, and either the smallest divides the order $3$ times, or the smallest prime divisor is $2$ (actually, by Feit-Thompson, we know this must be the case, but I preferred not to also invoke that), and $3$ must also divide the order.

This immediately gives us $60$ as a lower bound on the order of $G$, and the next possible order would be $2^2\cdot 3\cdot 7 = 84$ and all further orders are greater than $100$. So we are left with ruling out the order $84$ (which you have done), and showing that the only one of order $60$ is $A_5$.

Solution 2:

Regarding a group of order $24$.

We know that $n_2$ is either 1 or 3. If $n_2=1$ we are done. If $n_2=3$ then let $X$ be the set of the three $2$-Sylow groups. $G$ acts transitively on $X$ by conjugation. In particular, the action is not trivial. The group action can be thought of as a homomorphism $\phi\colon G\to S_X=S_3$. Since $1<\left|Im(\phi)\right|\le 6$ we have $8\le \left|\ker\phi\right|<24$, so $G$ is not simple.

You can significantly shorten the proof with the following lemma.

If $\left|G\right|=p^km$ where $p$ is a prime, $p\nmid m$ and $p^k\nmid (m-1)!$ then $G$ is not simple.

Proof: Let $P$ be a $p$-Sylow subgroup. Then $G$ acts transitively on the cosets $G/ P$ by left multiplication. Since $\left[G\colon P\right]=m$ this action is a homomorphism $\phi\colon G\to S_m$. If $G$ is simple then $\ker\phi=\{e\}$ so $\phi$ is an injection. That is, $G$ is isomorphic to a subgroup of $S_m$. Hence $p^km=\left|G\right| \mid m!$ so $p^k\mid (m-1)!$ which is a contradiction.

There is a small mistake in your argument, 36 is not a prime squared.

Solution 3:

Here's a possibly shorter way to do the problem from "scratch" (assuming only Sylow's theorem, and not assuming Burnside's $pq$-theorem) than what's done in OP:

Let $G$ be a simple nonabelian group of order $n < 100$. Let $f(n)$ be the largest prime dividing $n$.

Case-0: $n = p^k$. Then the center is nontrivial, and by cauchy on the center we can find a normal subgroup of order $p$. This is a proper subgroup when $k > 1$, and if $k = 1$ then $G$ is abelian. So assume $n$ is not a prime power. Thus if any sylow-$p$ group is normal, we will be done in the other cases (since that would be a proper normal subgroup)

Case-1: $f(n) > 7$. Then obviously $f(n) > 10$. Now, I claim the sylow-$f(n)$ group is normal. If it's not, then $n_{f(n)}(G) \geq 1 + f(n)$, so the group has order atleast $f(n)(f(n)+1) > 100$, a contradiction.

Case-2: $f(n) = 7$. Then $n_7(G) = 7k + 1$, and $7(7k+1) \leq 100 \Rightarrow n_7(G) \in \{1, 8 \}$. If $n_7(G) \neq 1$, then the $7 \cdot n_7 = 56$ divides the order of the group, which forces $n = 56$. If the sylow-$8$ group is not normal either, then $n_8 = 7$. Thus there are atleast $(7-1) \cdot n_7 + 1 + (8-1) + 1 > 56$ elements a in a group of order $56$, a contradiction.

Case-3: $f(n) = 5$. Using similar bounding, if the sylow-5 group is not normal, then $n_5(G) \in \{6, 11, 16 \}$.

If $n_5(G) = 16$, then $5 \cdot n_5 = 80 | n \leq 100 \Rightarrow n =80$. If the sylow-$16$ group is not normal, the $n_{16} = 5$. For two distnict sylow-16 groups $H_1, H_2$, note that $|H_1 \cap H_2| \leq 8$, so $|H_1 \cup H_2| = |H_1| + |H_2| - |H_1 \cap H_2| \geq 24$. Thus, there are exactly $(5-1) \cdot n_5 = 64$ elements of order exactly $5$, and atleast $24$ elements of order dividing $16$. But then $64 + 24 = 88 > 80$, a contradiction.
If $n_5(G) = 11$, then $11$ divides $n$ and this reduces to first case.
If $n_5(G) = 6$, then $30 | n$, so $n$ is either $30, 60, 90$. Now, note that a group of order $2k$ with $k$ odd has a normal subgroup of size $k$ (sketch: Let $G$ act on itself by left action, giving a faithful morphism $G \mapsto S_{|G|}$, compose with with the sign map $\pi \mapsto sgn(\pi)$ to get a map $G \mapsto \{1, -1 \}$. $G$ has an element of order $2$, and it's not hard to see this maps to $-1$, so the kernel of this map is the desired normal subgroup). So this gives the only possibility as $n = 60$

Case-4: Thus $n = 2^a 3^b$ for some $b$ with $a \geq 1$ (otherwise it's Case-0), and $1 \leq b \leq 4$).

$b = 4$. Then $n \geq 2 \cdot 3^4 = 162$, a contradiction.
$b = 3$. Then $n = 54 = 2 \cdot 27$, so it's handled by the $2k$ lemma (in $n_5(G) = 6$ case)
$b = 2$. Then $n = 2^k 9$ for $1 \leq k \leq 3$. Now, if $n_9 \neq 1$, then $n_9 = 10$ ($n_9 = 19$ is too large). But then $5 | n$, which reduces it to previous case.
$b = 1$. Then $n = 3 \cdot 2^k$.
- $n = 6$ is handled by the $2k$ lemma
- $n > 6$. Note that if $G$ is simple, and $H$ is a proper subgroup of $G$, then $G$ is isomorphic to a subgroup of $S_k$ where $k = [G:H] > 1$ (proof is easy, the action of $G$ on the left-cosets of $H$ by multiplication gives a map from $G$ to $S_k$. If the kernel is nontrivial, by simplicity it must be all of $G$, thus in particular $gH = H$ which is nonsense. Thus the kernel is trivial, and we conclude by first isomorphism). Thus in particular, $|G| \leq S_k = k!$ which (take $H$ to be sylow-$2^k$ group) is $6$ in this case. Again, a contradiction.