Isomorphism of multiplicative group modulo n Let's take groups $\mathbb{Z}^*_{12}, \mathbb{Z}^*_{10}$ and $\mathbb{Z}^*_{8}$ (multiplicative groups modulo  $12, 10$ and $8$).
The order of all these groups is $4$ (since $\varphi(12) = \varphi(10) =  \varphi(8) = 4$).
A multiplicative group modulo n with order of $4$ is known to be cyclic.
Therefore, there must exist an isomorphism between every one of these groups and the additive group $\mathbb{Z}_4^+$. Is that correct?
 A: The groups are
$$(\mathbb Z /12 \mathbb Z)^\times = \{1,5,7,11\}$$
$$(\mathbb Z /10 \mathbb Z)^\times = \{1,3,7,9\}$$
$$(\mathbb Z /8 \mathbb Z)^\times = \{1,3,5,7\}$$
As you can see all groups have the order 2 element -1 in them (11, 9, 7 respectively). ${{{}}}$
But $5^2 = 1$ and $7^2 = 1$ in the first group, whereas $3^2 = -1$ and $7^2 = -1$ in the second. That shows these two groups are not isomorphic.
A: Another approach to this is using the structure theorem of multiplicative group of units.
First by CRT you factor the groups into prime powers
$$(\mathbb Z /12 \mathbb Z)^\times = (\mathbb Z /2^2 \mathbb Z)^\times \times (\mathbb Z /3 \mathbb Z)^\times$$
$$(\mathbb Z /10 \mathbb Z)^\times = (\mathbb Z /2 \mathbb Z)^\times \times (\mathbb Z /5 \mathbb Z)^\times$$
$$(\mathbb Z /8 \mathbb Z)^\times = (\mathbb Z /2^3)^\times$$
Now use use the structure theorem which says


*

*$(\mathbb Z /2^r \mathbb Z)^\times = C_2 \times C_{2^{r-2}}$

*$(\mathbb Z /p^r \mathbb Z)^\times = C_{(p-1)p^{r-1}}$


this shows that first and last group are isomorphic (to $C_2 \times C_2$) while the middle one is different (it's $C_{2^2}$).
