Question is to Prove that :
$G$ is non abelian simple group of order $<100$ then $G\cong A_5$
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 🙂
Best Answer
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 $A_5$.