groups of order 112 and 120

fredbel6

there are no simple groups of order 112 and 120
(120 is tricky i guess because it contains a simple nonabelian group)

also in general, can two simple groups of same order be nonisomorphic?

**Posted:** Sun Dec 26, 2004 2:29 pm

jmerry

I don't know all of the simple groups off the top of my head, so I'm not sure about your first statement/question.

For your second, I have an example: There are two simple groups of order 168.

For the first, take the group of $3\times 3$ invertible matrices over the field with two elements. There are 168 of these, and they form a simple group under multiplication.
For the second, take the group of $2\times 2$ matrices of determinant 1 over the field with 7 elements. Take the quotient of this group by its center, $\pm I$ . This leaves a different simple group of order 168.

Both groups belong to the "PSL" family of simple groups.

**Posted:** Mon Dec 27, 2004 2:24 am

fredbel6

well my group theory professor said that there is a gag for group theory professors going on retirement : i am now as old as the order of the smallest nonabelian group, hopefully i don't reach that of the next one: and he meant A5 and L2( 7 )
and it is possible for a lot of orders like for instance 111,123,...
but what i meant was, is there a fast trick by using sylow groups and permutation representations on them etc.. to show that there aren't any of order 112 and 120
are they really nonisomorphic, those two groups of order 168?
i know that A5, L2(4) and L2(5 ) are all simple , but one can easily show that they all are isomorphic

thx for the help however

**Posted:** Mon Dec 27, 2004 7:45 am

jmerry

A proof that the two groups of order 168 are not isomorphic:

The first has no elements of order 8: in the base field $(x^8-1)=(x-1)^8$ , so a matrix of order dividing 8 has a power of $x-1$ as its minimal polynomial, and therefore has $(x-1)^3$ as its characteristic polynomial. Since $(x-1)^3|x^4-1$ , the matrix has order dividing 4.

In the second group, the equivalence class of $\begin{pmatrix}2&4 \\ 3&2 \end{pmatrix}$ has order 8:
$\begin{pmatrix}2 & 4 \\ 3& 2 \end{pmatrix}^2=\begin{pmatrix}2&2 \\ 5&2 \end{pmatrix}$
$\begin{pmatrix}2&4 \\ 3&2 \end{pmatrix}^4=\begin{pmatrix}0&1 \\ 6&0 \end{pmatrix}$
$\begin{pmatrix}2&4 \\ 3&2 \end{pmatrix}^8=\begin{pmatrix}1&0 \\ 0&1 \end{pmatrix}$

grobber

A really quick way to show that there are no simple non-abelian groups of order $112$ would be employing Burnside's Theorem: There are no simple non-abelian finite groups with order having less than $3$ distinct prime factors. Since $112=2^4\cdot 7$ , that's that.

I don't know how to prove the theorem, of course Smile

.

**Posted:** Mon Dec 27, 2004 7:03 pm

fredbel6

thank you jmerry for that explanation
grobber : that is an interesting theorem that gives me certainty about 112
( i hadn't heard about it yet Blush

)
of course if my professor would have allowed use of that, half of his problems
could have been discarded Mr. Green

so are there any people here who have a clue for 112

**Posted:** Tue Dec 28, 2004 5:42 am

fredbel6

hi

i'd like to come back to this topic

i see no hole at all in jmerry's proof

but on the other hand i have found a course claiming PSL(2,7) is isomorphic to PSL(3,2)

and on so many sites, like http://www.mathreference.com/grp-fin,g168.html

they speak of the simple group of order 168, as if it is unique up to isomorphism?

so can anyone explain what is going on, is there only one? or is the second one often forgotten?

thx

fred

**Posted:** Sun Feb 13, 2005 10:55 am

jmerry

They're wrong if they say it's the only one or that the two I described are isomorphic. The group described at the site you linked to is PSL(3,2); it has no elements of order 8 because it can be described as a permutation group on 7 elements. It's an oversight, and I had no idea it was common.
I guess people like to believe that groups must be isomorphic if you match a few simple properties (in this case, the number of Sylow subgroups of each order).

Here's a cute exercise to show that isomorphisms aren't always there when you think they should be: Find two non-isomorphic groups of order 16 which have the same number of elements of each order.

**Posted:** Sun Feb 13, 2005 5:31 pm

vess

A simple proof that no group of order $120$ is simple: Let $G$ be any group of order $120$ . By the Sylow theorems, the number of $5$ -Sylow subgroups of $G$ is either $1$ or $6$ ; in the first case, there is nothing to show because the unique Sylow $5$ -subgroup has to be normal. Thus, we may assume that $G$ has precisely $6$ Sylow $5$ -subgroups. Consider the action by conjugation of $G$ on the Sylow $5$ -subgroups; this action yields an embedding of $G$ into $S_6$ . As $G$ is simple, we must have $G \leq A_6$ , and $[A_6:G] = 3$ . This is a contradiction, because $A_6$ is simple and therefore cannot act on a three element set. Q.E.D.

**Posted:** Sun Feb 13, 2005 5:58 pm

fredbel6

okay so the overall tone here is that these groups are both simple with 168 elements but are not isomorphic

however i also asked this on 'ask an algebraist' (
http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist;task=show_msg;msg=1182.0001
)

this person claims it is actually a theorem in a book that they are isomorphic??

so what is going on, here people are certain that they aren't, and on other forums and in books the opposite is claimed?

sorry for insisting on this but i'm really interested

frederic

**Posted:** Tue Feb 15, 2005 1:55 pm

jmerry

I was about to post my proof there, but you did already. He's wrong, and the book being quoted is wrong; this short proof of nonisomorphism invalidates any proof of isomorphism, however long.

**Posted:** Tue Feb 15, 2005 2:05 pm

fredbel6

hi

on http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist;task=show_msg;msg=1182.0001.0001.0001

an argument against jmerry's proof is now given

the determinant of your representant is -1 which is no square modulo seven
it is, when we see PSL(2,7) as SL(2,7)*SC(2,7)/SC(2,7) , thus a subgroup of PGL(2,7)
only an element of PGL(2,7) and not of PSL(2,7)

so does this mean it is true that their is only one simple group of order 168

or can you fix your proof?

plz help,

fred

**Posted:** Wed Feb 16, 2005 10:27 am

jmerry

Now that I've had some time to think about it, it all falls apart. There are no elements of order 8 in either group; both have 2-Sylow subgroups isomorphic to the dihedral group.

In $PSL_3(F_2)$ , the upper triangular matrices form a 2-Sylow subgroup, which is easily seen to be the dihedral group.
In $PSL_2(F_7)$ , the "rotations" and "reflections" of the form $\begin{pmatrix}a&b\\-b&a\end{pmatrix}$ or $\begin{pmatrix}a&b\\b&-a\end{pmatrix}$ form a 2-Sylow subgroup. $\begin{pmatrix}2&2\\-2&2\end{pmatrix}$ and $\begin{pmatrix}2&-2\\2&2\end{pmatrix}$ are the elements of order 4, $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ is the non-identity element in the center, and the "reflections" $\begin{pmatrix}3&2\\2&-3\end{pmatrix}, \begin{pmatrix}3&5\\5&-3\end{pmatrix}, \begin{pmatrix}2&3\\3&-2\end{pmatrix}, \begin{pmatrix}2&4\\4&-2\end{pmatrix}$ are the other elements of order 2.

**Posted:** Wed Feb 16, 2005 1:58 pm

fredbel6

thanks for the remarks

i found it all very interesting

on this site i found a theorem :
http://www.maths.qmul.ac.uk/~pjc/class_gps/ch2.pdf

theorem 2.12 states that the PSL(2,p) with p odd prime are always unique simple groups of their order

so that implies PSL(2,7) being isomorphic with PSL(3,2)

thx

fred

**Posted:** Thu Feb 17, 2005 8:39 am

grobber

**Posted:** Thu Feb 17, 2005 11:52 pm