Let $G$ and $H$ be abelian groups. Show that the product $G\times H$ is also abelian.

Question

Let $G$ and $H$ be abelian groups. Show that the product $G\times H$ is also abelian.

I have already proved, that for groups $G$ and $H$ finite groups, then the direct product $G\times H$ is cyclical if and only if $G$ and $H$ are cyclical.

However, here I get confused because cyclic groups should be a subset of all abelian groups.

Answer 1

Let $(G,*)$ and $(H,\cdot)$ be abelian groups. Take $(a,b), (c,d) \in G\times H$ (write $+$ for the operation on $G\times H$). Now

$$\begin{align} (a,b)+(c,d) &= (a*c, b\cdot d) \\ &= (c*a, d\cdot b)\\ & = (c,d)+(a,b). \end{align}$$

This shows $(G\times H, +)$ is abelian.

Answer 2

Welcome to MSE!

Hint:

Do you remember how multiplication is defined in $G \times H$?

Fix two elements $(g_1,h_1)$ and $(g_2,h_2)$ in the direct product.

Can you do a simple computation to show $(g_1,h_1)(g_2,h_2) = (g_2,h_2)(g_1,h_1)$?

You'll need the fact that $G$ and $H$ are both abelian here.

---

I hope this helps ^_^

Answer 3

Let $(g_1, h_1)$ and $(g_2, h_2)$ be two elements from $G \times H$. Use the definition of the group operation in $G \times H$ to write down what $(g_1, h_1) (g_2, h_2)$ is. Do the same for $(g_2, h_2) (g_1, h_1)$. Then use the fact that $G$ and $H$ are abelian to show that these two elements are the same.