While practising for an exam I encountered the following question.
Consider the groups $G_1=(\mathbb{Z}/100\mathbb{Z})^{\times}$ and $G_2=(\mathbb{Z}/110\mathbb{Z})^{\times}$ and $G_3=(\mathbb{Z}/132\mathbb{Z})^{\times}$.
(a) Show that these three groups have the same number of elements.
(b) Show that $G_1$ and $G_3$ are not isomorphic.
(c) Show that $G_1$ and $G_2$ are isomorphic. (Hint: probably the shortest way to do this, is to use both the Chinese Remainder Theorem and the theory of elementary divisors).
I know how to solve $(a)$, we use Euler's totient function and find that they all have 40 elements. However at $(b)$ I get stuck. I tried prime factorisation: $$100 = 2 \cdot 2 \cdot 5 \cdot 5\\
132 = 2 \cdot 2 \cdot 3 \cdot 11 $$
Then by the chinese remainder theorem we find
$$G_{1}=(\mathbb{Z} / 100 \mathbb{Z})^{x} \cong (\mathbb{Z} / 2 \mathbb{Z})^{\times} \times (\mathbb{Z} / 2 \mathbb{Z})^{\times} \times (\mathbb{Z} / 5 \mathbb{Z})^{\times} \times (\mathbb{Z} / 5 \mathbb{Z})^{\times} \\G_{3}=(\mathbb{Z} / 132 \mathbb{Z})^{x} \cong (\mathbb{Z} / 2 \mathbb{Z})^{\times} \times (\mathbb{Z} / 2 \mathbb{Z})^{\times} \times (\mathbb{Z} / 3 \mathbb{Z})^{\times} \times (\mathbb{Z} / 11 \mathbb{Z})^{\times}$$ Now here I am stuck. Any help would be appreciated. I assume $(c)$ can be done in a similar way?
Best Answer
You applied the chinese remainder theorem incorrectly. Recall that if $n = p_1^{\alpha_1}\ldots p_k^{\alpha_k}$ the chinese remainder theorem states that $\mathbb Z/n\mathbb Z \cong \mathbb Z/p_1^{\alpha_1}\mathbb Z \times \cdots \times \mathbb Z/p_k^{\alpha_k}\mathbb Z$ (and similarly for the multiplicative group).
Thus, we have $$(\mathbb Z/100\mathbb Z)^* \cong (\mathbb Z/4 \mathbb Z)^* \times (\mathbb Z/25 \mathbb Z)^*.$$
Note that $(\mathbb Z/4\mathbb Z)^*\cong C_2$ (where $C_n = \mathbb Z/n\mathbb Z$ for brevity) and $(\mathbb Z/25 \mathbb Z)^* \cong C_{20}$, hence applying the chinese remainder theorem again we find $G_1 = (\mathbb Z/100 \mathbb Z)^* \cong C_2 \times C_{20} \cong C_2 \times C_4 \times C_5$.
Similarly, we have $G_2 = (\mathbb Z/110\mathbb Z)^* \cong (\mathbb Z/2\mathbb Z)^* \times (\mathbb Z/5\mathbb Z)^* \times (\mathbb Z/11\mathbb Z)^* \cong C_4 \times C_{10} \cong C_4 \times C_2 \times C_5$
$G_3 = (\mathbb Z/132\mathbb Z)^* \cong (\mathbb Z/4\mathbb Z)^* \times (\mathbb Z/3\mathbb Z)^* \times (\mathbb Z/11\mathbb Z)^* \cong C_2 \times C_2 \times C_{10} \cong C_2 \times C_2 \times C_2 \times C_5.$
It's then clear that $G_1 \cong G_2$, and showing that $G_1 \not\cong G_3$ should be straightforward.