Hw9 Problem 4

### Problem 4

Mark the following as true or false.
(a) $\mathbb Z_8$ is generated by {4,6}
(b) Every abelian group of order divisible by 5 contains a cyclic subgroup of order 3.
(c) $\mathbb Z_2\times\mathbb Z_4$ is isomorphic to $\mathbb Z_8$.
(d) $\mathbb Z_2\times\mathbb Z_4$ is isomorphic to $S_8$.
(e) The order of $\mathbb Z_{12}\times\mathbb Z_{15}$ is 60.

Solution
(a) False. Because 4 and 6 are not relatively prime, so there's no way to write 1 as a sum of 4's and 6's.
(b) False. The subgroup must divide the order of the group.
(c) False. $\mathbb{Z}_8$ is cyclic, but $\mathbb{Z}_{2}\times \mathbb{Z}_{4}$ is not.
(d) False. $\mathbb{Z}_{2}\times \mathbb{Z}_{4}$ is abelian, but $S_8$ is not.
(e) False. The order of $\mathbb{Z}_{12} \times \mathbb{Z}_{15}$ is $12*15=180$