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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?

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Are you sure that any group of order $4$ is cyclic ? how about $\frac{\mathbb{Z}}{2\mathbb{Z}}\times \frac{\mathbb{Z}}{2\mathbb{Z}}$ ? –  Louis La Brocante Jan 7 '13 at 11:28
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It is at least abelian, but not always cyclic. –  André Jan 7 '13 at 11:30
    
Find an element of order $4$ in the group @Rolando gave. –  B. S. Jan 7 '13 at 11:35
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2 Answers

up vote 2 down vote accepted

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 $C_{2^2}$).

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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, 8 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.

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