What are the subgroups of $\mathbb{Z}_{5}^\ast \times \mathbb{Z}_{4}^\ast$? I think I count $5$ of them. What are the subgroups of $\mathbb{Z}_{5}^\ast \times \mathbb{Z}_{4}^\ast$? I think I count $5$ of them I think? Thank you for anyone that puts the time in to help!
 A: As @Ian posted nicely $$\mathbb Z_5^*\times\mathbb Z_4^*\cong\mathbb Z_4\times\mathbb Z_2$$ Now think about the latter group. If $H\leq \mathbb Z_4\times\mathbb Z_2=G$ then $|H|=1,~2,~4,~\text{or}~~8$ according to Lagrange's theorem and this fact that $|\mathbb Z_4\times\mathbb Z_2|=8$. Since we are working with easy groups so we can list all elements of $G$ as follows:
$$(0_{\mathbb Z_4},0_{\mathbb Z_2}),~~(1_{\mathbb Z_4},0_{\mathbb Z_2}),~~(2_{\mathbb Z_4},0_{\mathbb Z_2}),~~(3_{\mathbb Z_4},0_{\mathbb Z_2})\\
(0_{\mathbb Z_4},1_{\mathbb Z_2}),~~ (1_{\mathbb Z_4},1_{\mathbb Z_2}),~~(2_{\mathbb Z_4},1_{\mathbb Z_2}),~~(3_{\mathbb Z_4},1_{\mathbb Z_2})$$ Clearly some subgroups of $G$ is the cyclic ones, but we should care about this point that some of them may be generated twice. For example: $$H_1=\langle(0,0)\rangle=\langle0_G\rangle\\H_2=\langle(1,0)\rangle=\{(1,0),(2,0),(3,0),(0,0)\}\cong\mathbb Z_4\\ H_3=\langle(2,0)\rangle=\{(2,0),(0,0)\}\cong\mathbb Z_2\\ H_4=\langle(3,0)\rangle=\{(3,0),(2,0),(1,0),(0,0)\}\cong\mathbb Z_4\\ H_5=\langle(0,1)\rangle=\{(0,1),(0,0)\}\cong\mathbb Z_2 $$  Using GAP (in particular, AllSubgroups and StructureDescription), possible isomorphism types of subgroups of $G$ are:
1, C2, C2 x C2, C4, C4 x C2

A: $(\mathbb Z/5\mathbb Z)^\times\cong\mathbb Z/4\mathbb Z$. $(\mathbb Z/4\mathbb Z)^\times\cong \mathbb Z/2\mathbb Z$.
