Since the task requires elements of order 12 in a direct product group, we look for elements $a,b,c$ such that $lcm(|a|,|b|,|c|) = 12$, and $a \in \mathbb{Z}_3, b \in \mathbb{Z}_4, c \in U(9)$. Possible orders in $\mathbb{Z}_3$ are $1,3$; in $\mathbb{Z}_4$ are $1,2,4$; and in $U(9)$ are $1,2,3,6$ (I calculated these). Thus for the elements of order 12 we have the following list of possible orders: $(1,4,3), (1,4,6),(3,4,1),(3,4,3),(3,4,6),(3,4,2)$ and therefore, we calculate the phi-Euler totient function (ex.$\phi(1)\phi(4)\phi(3)$) for all of them, and add them up. My answer was 28 but the total in the exercise is 32.
Did I miss something or is my way of solving not correct?Thanks for any help.
Update: I should practice my multiplication skills.
I obtain $4+4+4+8+8+4=32$ for the sum of the six pairs: $$ \phi(1)\phi(4)\phi(3)=4,\; \phi(1)\phi(4)\phi(6)=4,\;\phi(3)\phi(4)\phi(1)=4, $$
$$ \phi(3)\phi(4)\phi(3)=8,\; \phi(3)\phi(4)\phi(6)=8,\;\phi(3)\phi(4)\phi(2)=4. $$ So I suppose you computed one of the values differently.