This is not a complete answer in any sense, but I will make a few comments. The irreducible subgroups $G$ of ${\rm GL}(2,\mathbb{C})$ are the primitive ones, which have $G/Z(G)$ isomorphic to $A_{4},S_{4}$ or $A_{5}$, and imprimitive groups, which have an Abelian normal subgroup of index $2.$ On the other hand, any finite group with an Abelian normal subgroup of index $2$ has all its irreducible representations of degree $1$ or $2,$ so the number of $2$-dimensional irreducible representations is easily calculated. A more careful analysis of the primitive case shows that if $G$ has a faithful $2$-dimensional primitive complex representations, then $G = Z(G)E,$ where $ E \cong {\rm SL}(2,3), {\rm GL}(2,3),{\rm SL}(2,5)$ or the binary icosahedral group (also, a double cover of order $48$ of $S_{4},$ (as ${\rm GL}(2,3)$ is), but with a generalized quaternion Sylow $2$-subgroup). Now let $G$ be any finite group, and let $K$ be the intersection of the kernels of the irreducble complex representation of $G$ of degree at most $2$. The above discussion means that the only possible non-Abelian composition factor of $G/K$ is $A_{5}$, though it may be repeated if it appears. The answer to your question only depends on the structure of $G/K,$ so we may reduce to the case that all composition factors of $G$ are cyclic or $A_{5}.$ Also, by Clifford theory, we may suppose that the Fitting subgroup $F(G)$ is a direct product of an Abelian group of odd order and a $2$-group, and that all components of $G$ (if there are any) are isomorphic to ${\rm SL}(2,5).$ Further analysis can be carried out, but I believe that the analysis is not entirely straightforward. Perhaps this outline will help others to complete it, so I submit it.
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This is not a complete answer in any sense, but I will make a few comments. The irreducible subgroups $G$ of ${\rm GL}(2,\mathbb{C})$ are the primitive ones, which have $G/Z(G)$ isomorphic to $A_{4}$ A_{4},S_{4}$ or $A_{5}$, and imprimitive groups, which have an Abelian normal subgroup of index $2.$ On the other hand, any finite group with an Abelian normal subgroup of index $2$ has all its irreducible representations of degree $1$ or $2,$ so the number of$2$-dimensional of $2$-dimensional irreducible representations is easily calculated. A more careful analysis of the primitive case shows that if $G$ has a faithful $2$-dimensional primitive complex representations, then $G = Z(G)E,$ where $ E \cong {\rm SL}(2,3)$ SL}(2,3), {\rm GL}(2,3),{\rm SL}(2,5)$ or the binary icosahedral group (also, a double cover of order $48$ of $S_{4},$ (as ${\rm SL}(2,5).$ GL}(2,3)$ is), but with a generalized quaternion Sylow $2$-subgroup). Now let $G$ be any finite group, and let $K$ be the intersection of the kernels of the irreducble complex representation of $G$ of degree at most $2$. The above discussion means that the only non-Abelian composition factor of $G/K$ is $A_{5}$, though it may be repeated. The answer to your question only depends on the structure of $G/K,$ so we may reduce to the case that all composition factors of $G$ are cyclic or $A_{5}.$ Also, by Clifford theory, we may suppose that the Fitting subgroup $F(G)$ is a direct product of an Abelian group of odd order and a $2$-group, and that all components of $G$ (if there are any) are isomorphic to ${\rm SL}(2,5).$ Further analysis can be carried out, but I believe that the analysis is not entirely straightforward. Perhaps this outline will help others to complete it, so I submit it. |
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This is not a complete answer in any sense, but I will make a few comments. The irreducible subgroups $G$ of ${\rm GL}(2,\mathbb{C})$ are the primitive ones, which have $G/Z(G)$ isomorphic to $A_{4}$ or $A_{5}$, and imprimitive groups, which have an Abelian normal subgroup of index $2.$ On the other hand, any finite group with an Abelian normal subgroup of index $2$ has all its irreducible representations of degree $1$ or $2,$ so the number of$2$-dimensional irreducible representations is easily calculated. A more careful analysis of the primitive case shows that if $G$ has a faithful $2$-dimensional primitive complex representations, then $G = Z(G)E,$ where $ E \cong {\rm SL}(2,3)$ or ${\rm SL}(2,5).$ Nowlet Now let $G$ be any finite group, and let $K$ be the intersection of the kernels of the irreducble complex representation of $$ G$ of degree at most $2$. The above discussion means that the only non-Abelian composition factor of $G/K$ is $A_{5}$, though it may be repeated. The answer to your question only depends on the structure of $G/K,$ so we may reduce to the case that all composition factors of $G$ are cyclic or $A_{5}.$ Also, by Clifford theory, we may suppose that the Fitting subgroup $F(G)$ is a direct product of an Abelian group of odd order and a $2$-group, and that all components of $G$ (if there are any) are isomorphic to ${\rm SL}(2,5).$ Further analysis can be carried out, but I believe that the analysis is not entirely straightforward. Perhaps this outline will help others to complete it, so I submit it. |
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