I'm trying to show that $Z\_2 \times Z\_2 \times Z\_2$ is isomorphic to $U(24)$.

I start by defining $f \colon Z\_2 \times Z\_2 \times Z\_2 \to U(24)$ by $f((a,b,c)) = 12a + 6b + 4c + 1 \pmod{24}$.

Let's attempt to show it preserves the operation:

$$ \begin{align\*}
f((a,b,c))f((d,e,f)) &= 
(12a+6b+4c+ 1)(12d+6e+4f+1) \pmod{24} \\ &= 
144ad + 72ae + 48af + 12a + 72bd + 36be + 24bf + 6b \\ & \quad + 48cd + 24ec + 16cf + 4c + 12d + 6e + 4f + 1 \pmod{24} \\ &= 
12a + 12d + 6b + 6e + 4c + 4f + 1 + 36be + 16cf \pmod{24} \\ &= f((a+d,b+e,c+f)) + 36be + 16cf \pmod{24} \\ &= 
f((a,b,c)(d,e,f)) + 36be + 16cf \pmod{24}
\end{align\*} $$

It seems like the extra $36be$ and $16cf$ terms are saying that $f$ does not preserve the operation...

However, I have been unable to find a counter example of the map not working (not a proof obviously, but I'm just wondering where I'm going wrong).