Product of cyclic groups

Question

How can you quickly tell that the product of cyclic groups $\mathbb{Z}_4 \times \mathbb{Z}_3$ has a 2-subgroup containing an element of order 4? Also, I don't understand the notion of multiplying groups like this. If a Cartesian product is an ordered pair, is this product the group of all ordered pairs of cyclic subgroups from $\mathbb{Z}_4$ and $\mathbb{Z}_3$? Do you have to manually write out all the elements of the product in order to see that there is a 2-subgroup containing an element of order 4, or is there a faster method besides memorizing it?

Answer 1

Things get a bit more complicated when you look at things more generally. If you write groups with identity equal to $1$ then $G \times H$ always has a subgroup isomorphic to $G$ consisting of the elements of form $(g,1)$ in the product. You should also be able to find a group isomorphic to $H$.

In your group, there is also an element $z$ of order $12$ (which may or may not be obvious to you). If you know this, then $z^3$ has order $4$.

Answer 2

Lemma: Let $\mathbb Z_n$ be cyclic group of order $n$. Then $\mathbb Z_n\times \mathbb Z_m \cong \mathbb Z_{nm}$ where $(m,n)=1$.

By previous lemma, one can see $\mathbb Z_3\times \mathbb Z_4 \cong \mathbb Z_{12}$. Since $\mathbb Z_{12}$ is cyclic, there are exactly one $2$-sylow subgroup-it is isomorphic to $\mathbb Z_4$. Hence, the element of order $4$ is contained in the $2$-sylow subgroup.