Counting Subgroups

Since $4036 = 2^2 \times 1009$ , the ring $\mathbb{Z}_{4036}$ is isomorphic to the direct product of rings $\mathbb{Z}_4 \times \mathbb{Z}_{1009}$ , and hence the group of units $G = \mathbb{Z}_{4036}^\times$ is isomorphic to the direct product of cyclic groups $\mathbb{Z}_4^\times \times \mathbb{Z}_{1009}^\times \; \equiv \; C_2 \times C_{1008}$ Let us determine the number of subgroups of $C_2 \times C_N$ . Suppose that $C_2$ is generated by $a$ . Let $f:C_2 \times C_N \to C_N$ be the projection homomorphism. If $H$ is a subgroup of $C_2 \times C_N$ , then $f(H)$ is a subgroup of $C_N$ , and so is cyclic, generated by some $c \in C_N$ . Let $S = \{x \in C_2 \,|\, (x,c) \in H\}$ . There are 3 cases to consider:

• If $S = \{e\}$ , then $H = \{e\}\times f(H)$ .
• If $S = C_2$ , then $H = C_2 \times f(H)$ .
• If $S = \{a\}$ , suppose that $o(c )$ was odd. Then $(a,e) \; = \; (a,c)^{o(c )} \in H$ , which would imply that $(e,c) = (a,e)(a,c) \in H$ , and hence $e \in S$ . Thus, in this case, we must have $o( c)$ being even, in which case it follows that $H$ is the cyclic group generated by $(a,c)$ .

We deduce that $C_2 \times C_N$ has two subgroups for each subgroup of $C_N$ of odd order, and three subgroups for each subgroup of $C_N$ of even order. Now $C_N$ contains one subgroup for each positive divisor of $N$ . Thus if we write $N = 2^u M$ where $u \ge 0$ is an integer and $M$ is an odd integer, then $C_2 \times C_N$ possesses $3\sigma_0(N) - \sigma_0(M)$ subgroups, where $\sigma_0(n)$ is the number of positive integer divisors of $n$ .

Finally, since $1008 = 2^4 \times 3^2 \times 7$ we deduce that $G = \mathbb{Z}_{4036}^\times$ has $3 \times \sigma_0(1008) - \sigma_0(63) \; = \; 3(5\times3\times2) - (3\times2) \; = \; 84$ subgroups.