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it is well known that the subgroups of SL(2,C) can be determined, I am wondering if the same situation is known for SL(4,C).For example, I want to know if group of order 42 can be a subgroup of SL(4,C).Thanks for consideration.

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$SL(2,\mathbb{C})$ contains cyclic groups of all orders, therefore $SL(k,\mathbb{C})$ for $k\geq 2$ does as well. Maybe you wanted to ask something different, e.g. if $SL(4,\mathbb{C})$ contains subgroups of order $42$ of all isomorphy types? –  Johannes Hahn Apr 15 '12 at 17:04
Certainly you have an embedding $SU(1)$ into $SL(4,C)$. –  plusepsilon.de Apr 15 '12 at 17:05
see Eric Rowell's answer to question mathoverflow.net/questions/17072 –  Bruce Westbury Apr 15 '12 at 17:05

1 Answer 1

Any group of order 42 is isomorphic to $C_{42}$, $AGL(1,7) = C_{7} \rtimes C_{6}$, $D_{14} \times C_{3}$, $D_{6} \times C_{7}$, $D_{42}$ or $(C_{7} \rtimes C_{3}) \times C_{2}$. (Here $D_{2n}$ denotes the dihedral group of order $2n$.)
Any finite cyclic group can be embedded in $GL(3, \mathbb{C})$ as a group of scalar matrices, so those factors in the Cartesian products given can be taken care of. The largest dimension of an irreducible (complex) representation of a dihedral group is 2, and the largest dimension of an irreducible representation of $C_{7} \rtimes C_{3}$ is 3. These largest-dimensional representations are all faithful. So these groups are all subgroups of $GL(3, \mathbb{C})$, which is isomorphic to a subgroup of $SL(4, \mathbb{C})$.
This leaves only $C_{7} \rtimes C_{6}$. The irreducible representations of this group have dimensions 1, 1, 1, 1, 1, 1 and 6. Any combination of the 1-dimensional representations will be an abelian representation and thus not be faithful. The 6-dimensional representation can't be found in $SL(4, \mathbb{C})$, so any group of order 42 will be isomorphic to a subgroup of $SL(4, \mathbb{C})$ or isomorphic to $AGL(1,7) = C_{7} \rtimes C_{6}$.

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