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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 :)

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    $\begingroup$ Wow! +1 for such a well thought out question! I think it must be the most detailed question with the most research effort shown that I've seen on this website! $\endgroup$ – Amitesh Datta Aug 31 '13 at 11:26
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    $\begingroup$ Agree with Amitesh, but it may make someone exhausted. It needs good revisions. :) $\endgroup$ – Mikasa Aug 31 '13 at 11:30
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    $\begingroup$ @AmiteshDatta : I was actually thought of leaving it question partially answered i.e., to the case of groups of prime orders.But then i tried some thing and it worked out :) $\endgroup$ – user87543 Aug 31 '13 at 11:32
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    $\begingroup$ A general hint: If a simple group $G$ has a subgroup $H$ of index $n$, then its action on the cosets of $H$ gives rise to a homomorphism for $G\to S_n$. As $G$ is simple, this homomorphism must be injective, so $G$ is a subgroup of $S_n$. Many cases can be eliminated, if you know the simple groups that occur as subgroups of $S_4$, $S_5$,... $\endgroup$ – Jyrki Lahtonen Aug 31 '13 at 11:43
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    $\begingroup$ Does $96$ divide $3!$? Also, I should state that group actions are perhaps one of the most important aspects of group theory. An excellent way to understand a group is to understand the sets on which it acts. E.g., @Jyrki has provided a fine example of this in the context of your question but there are so many other examples. For example, from geometric topology where one can think of a group as acting on a smooth manifold and then one can use the topology of the manifold to understand the group. (I mention this because I'm thinking about this kind of thing at the moment!) Cheers, $\endgroup$ – Amitesh Datta Sep 1 '13 at 12:40
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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 $A_5$.

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  • $\begingroup$ I am not supposed to use burnside theorem, I have learnt that in my "Representation theory of finite groups " course which is much after group tehory course. I am assuming i have nothing else than group actions, sylows theorems.. But, I liked it. Thank you!! $\endgroup$ – user87543 Sep 5 '13 at 12:40
  • $\begingroup$ I follow the argument that the smallest prime has to divide the order of a nonabelian simple group at least twice, but I am missing something about the $p^3$-or-$12$ argument. Why can't $N_G(P)/C_G(P)$ have order divisible by $p$? ... Oh I see why. We have to invoke the fact that if $|P|=p^2$ then $P$ is abelian, so that $C_G(P)\supset P$, and then $N_G(P)/C_G(P)$ can't be divisible by $p$ without implying that $P$ is not really $p$-Sylow. $\endgroup$ – Ben Blum-Smith Jun 23 '17 at 14:01
  • $\begingroup$ (Also, just to make explicit where the $12$ comes from, this happens because when $p=2$, $p+1$ is a prime so the case that $|P|=p^2$ and $P$ is not cyclic, so that $N_G(P)/C_G(P)$ has order $p+1=3$, becomes possible. And indeed, the 2-sylow of $A_5$ is not cyclic.) $\endgroup$ – Ben Blum-Smith Jun 23 '17 at 14:05
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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.

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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.
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