# Maximal Subgroups cannot be generated.

Quick question. I have a list of conjugacy classes (in ATLAS notation) such as:

(2A, 3A, 8A) and (2A, 4A and 6A) - these are triples of conjugacy classes that are present in $M_{12}$ (Mathieu Group). I wish to show that these triple of elements do not generate $M_{12}$, but rather they generate proper subgroups of $M_{12}$.

EDIT: I'm extremely sorry, I forgot to add the condition that the for the conjugacy classes $(C_1, C_2, C_3)$ of $M_{12}$ we require that for $g_i \in C_i$, $M_{12}$ = $<g_1, g_2, g_3>$ with $g_1g_2g_3 = 1$ (that the product of elements from each conjugacy class in the triple is the identity element of the group).

Apparently, one can retrieve this information by looking at the character tables of the maximal subgroups of $M_{12}$ (as one can do so in GAP) to show that $M_{12}$ cannot be generated by the above conjugacy classes. My question is essentially how do the character tables of the maximal subgroups help us see that the triple of classes do not generate the entire group, $M_{12}$.

Best,

Always

• A subgroup is normal iff it is the union of conjugacy classes. Since $M_{12}$ is simple, any nontrivial conjugacy class generates the whole group. Am I interpreting your question correctly? – Josh B. Aug 31 '17 at 0:04
• Hi, thank you ever so much for your comment. I'm extremely sorry - it was late at night and I forgot to add an important condition that the generators must satisfy. I believe that the condition may have changed now (please see original post that has now been edited). – AlwaysNeedHelp Aug 31 '17 at 8:23

I looked at this, and to me it seems that what you want to show is false.

Using the online ATLAS, you can construct $G = M_{12}$ with its standard generators $a, b$ in MAGMA or GAP. The ATLAS will also give you representatives for each conjugacy class of $G$ as a word in $a$ and $b$. Thus you can construct representatives $x$, $y$, $z$ for the conjugacy classes $2A$, $4A$, $6A$, say.

With a random search, you will find that most of the time, for $g \in G$ and $h \in H$ the subgroup $\langle x, gyg^{-1}, hzh^{-1} \rangle$ is all of $G$. The same is also true for the triple $(2A, 4A, 8A)$ of conjugacy classes.

EDIT: The above answers the original question. For the modified question, there are indeed character theoretic methods. See for example the following paper, especially pg. 31-32.

Woldar, Andrew J.; Representing $M_{11}$, $M_{12}$, $M_{22}$ and $M_{23}$ on surfaces of least genus. Comm. Algebra 18 (1990), no. 1, 15–86.

The key point is that if $C_1$, $C_2$, $C_3$ are conjugacy classes of $G$, then from the character table of $G$ one can calculate the number of triples $(g_1, g_2, g_3)$ such that $g_i \in C_i$ and $g_1g_2g_3 = 1$.

There are also other methods, for example:

Conder, Marston; The symmetric genus of the Mathieu groups. Bull. London Math. Soc. 23 (1991), no. 5, 445–453.

• Hi, thank you ever so much for your comment. I'm extremely sorry - it was late at night and I forgot to add an important condition that the generators must satisfy. I believe that the condition may have changed now (please see original post that has now been edited). – AlwaysNeedHelp Aug 31 '17 at 8:23
• @AlwaysNeedHelp: See the edit. There is a lot of literature dealing with these kinds of generation problems. – Mikko Korhonen Aug 31 '17 at 9:18