Classify all finite rings such that each unit has order 24 Problem: Suppose $R$ is a finite (associative) ring with 1 such that every unit of $R$ has order dividing 24. Classify all such $R$. 
My attempt: I had to quotient out the jacobson radical $J(R)$ so that since $R$ is finite and hence artinian, $R/J(R)$ is semisimple. Hence, $R/J(R)$ is a product of matrix rings, and by analyzing their general linear groups (and using the fact that all the units have order dividing 24) I am able to classify all such $R/J(R)$. 
But now I am stuck. I have no clue how to go from $R/J(R)$ to $R$. I have no information about $J(R)$. My brainstorming has this to say: Since $R$ is artinian, $J(R)$ is the double sided ideal of all nilpotent elements. Also, $\forall x \in J(R)$, $1+x$ is a unit. This defines an injective map from $J(R)$ to the set of units in $R$. But this is not a group homomorphism, so I cannot take advantage of the fact that the units have order dividing $24$.
 A: By the way, if I were to assume semisimplicity for $R$ (which is what $R/J(R)$ is in the original question), below is my solution:
Since $R$ is finite and hence artinian, Artin-Wedderburn implies that $R \cong M_{n_1}(D_1) \times \cdots \times M_{n_t}(D_t)$, where $n_i$ are positive integers and $D_i$ are division rings. Thus the group of units of $R$ is isomorphic to $GL_{n_1}(D_1) \oplus \cdots \oplus GL_{n_t}(D_t)$. Since $R$ is finite, so is each $D_i$, so that by Wedderburn's little theorem, each $D_i$ is in fact a finite field. Every unit of $R$ has multiplicative order dividing 24, and since $\forall \alpha \in D_i$, $\alpha I_{n_i} \in GL_{n_i}(D_i)$, we have that each nonzero element of $D_i$ has order dividing 24. Since $D_i$ is a finite field, the multiplicative group of units of $D_i$ must be cyclic. Therefore, the order of the group must divide 24. Thus, $|D_i|$ is of the form $k+1$, where $k|24$. Since $D_i$ is a finite field, this $k+1$ is necessarily a prime power. We then check all the divisors of 24, and it turns out that for each $k|24$, $k$ is a prime power, and we get that $D_i$ can be a finite field of order 2,3,4,5,7,9,13, or 25. Letting $q_i = |D_i|$, we then have that $|GL_{n_i}(D_i)| = (q_i^{n_1} - 1) \cdots (q_i^{n_i} - q_i^{n_i - 1}) = q_i^{n_i \choose 2} \prod_{j=1}^{n_i}(q_i^j - 1)$. Since each element of $GL_{n_i}(D_i)$ has order dividing $24 = 2^3\cdot3$, and since (by Cauchy) for every prime divisor $p$ of $|GL_{n_i}(D_i)|$ there exists an element in the group of order $p$, we must have that $|GL_{n_i}(D_i)|$ is a product of powers of 2 and 3. Using this and the observation that $|GL_n(D_i)|$ divides $|GL_{n+1}(D_i)|$ for each $n$, we conclude that $n_i$ is at most 2 when $q_i \in \{2,3\}$ and $n_i = 1$ otherwise. If $q_i = 2$, then $|GL_2(D_i)| = 2\cdot3$ and everything checks out. If $q_i = 3$, then $|GL_2(D_i)| = 2^4\cdot3$, and so in this case $n_i$ can be 2 provided that the Sylow-2 subgroup of $GL_2(D_i)$ is not cyclic. Now $GL_2(D_i)$ has 2 distinct elements of order 2, and so its Sylow-2 subgroup cannot be cyclic, and therefore it is a possibility. 
To summarize, we thus have $R \cong F_1 \times \cdots \times F_s \times M_2(G_1) \times \cdots \times M_2(G_t)$, where $s,t$ are integers, $F_i$ are fields of order 2,3,4,5,7,9,13, or 25 and $G_i$ are fields of order 2 or 3.
