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Original problem: suppose a group $G$ has irreducible complex representations of dimensions $1,1,2,3$, and $d$. Find $d$.

Using some basic dimension counting, we immediately get $d=3$ or $d=15$. The goal is to eliminate $d=15$.

If $d=15$, then $|G|=1^2+1^2+2^2+3^2+15^2=240$. Since $G$ has two $1$-dimensional irreducible representations, $[G:G']=2$, and $G'$ is normal. Thus $G'$ is a union of conjugacy classes. But $G'$ has elements of order $1$, $2$, $3$, and $5$, so it contains at least $4$ conjugacy classes. The only possibility is that $G\setminus G'$ is a single conjugacy class of order $120$.

At this point, it is driving me crazy that I can't find an easier contradiction. It seems absurd that $G$ can have $240$ elements with half of them in a single conjugacy class. I've included a solution I found which I am comfortable with, but which seems a bit convoluted to me.

Solution: For any $x\in G\setminus G'$, the centralizer $C_G(x)$ has order $2$, so $x^2=1$. Then $C_G(x)$ extends to a Sylow $2$ subgroup $P$. The center of $P$, $Z(P)$, is nontrivial, and $Z(P)\subseteq C_G(x)$, so $C_G(x)=Z(P)$. But if $x$ is central in $P$, then $|C_G(x)|\ge |P|=16$.

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    $\begingroup$ I actually quite like that argument. It generalizes nicely to rule out centralizers of any prime order when the square of that prime divides the order of the group, and this can be useful to know in many cases. $\endgroup$
    – Tobias Kildetoft
    Jul 31, 2018 at 5:57
  • $\begingroup$ Another proof, although it turns out to probably be worse: the group $G'$ must be a sum of the size of $4$ conjugacy classes, one of which has size $1$. There is no way to write $119$ using three divisors of $240$. $\endgroup$
    – pancini
    Jul 31, 2018 at 6:14
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    $\begingroup$ That is also fine, but I like your first one as that gives a more general approach that can also be used for other problems. $\endgroup$
    – Tobias Kildetoft
    Jul 31, 2018 at 6:15
  • $\begingroup$ I had hoped to give a solution using only character table properties, but made a slip in the solution I posted and have now deleted: sorry. There is a possible array satisfying all the usual orthogonality properties. $\endgroup$
    – ancient mathematician
    Aug 1, 2018 at 6:49

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A different approach is to use a list of all $8$ groups with $5$ conjugacy classes. According to GAP the orders of these groups are $\ 5,8,14,20,21,24,60. \ $ Because the order of the group must be $\ n := 1^2+1^2+2^2+3^2+d^2 = 15+d^2 \ $ and we can eliminate groups with prime order, that leaves $\ 16,24,40,51,64,\dots \ $ and so that implies $\ n = 24 \ $ and GAP has only $\ S_4 \ $ in the first list.

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