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Problem 98 (External Direct Products Of Abelian And Cyclic Groups)

Suppose that $G$ and $H$ are groups.

  1. Prove or disprove: If both $G$ and $H$ are Abelian, then $G\oplus H$ is Abelian.
  2. Prove or disprove: If $G\oplus H$ is Abelian,then both $G$ and $H$ are Abelian.
  3. Prove or disprove: If both $G$ and $H$ are cyclic, then $G\oplus H$ is cyclic.
  4. Prove or disprove: If $G\oplus H$ is cyclic, then both $G$ and $H$ are cyclic.
Hint: One of the four statements above is false. The rest are all true.

Solution

We first prove that if both $G$ and $H$ are abelian, then $G\oplus H$ is abelian. Suppose $G$ and $H$ are abelian. Pick $a,b\in G$. Pick $c,d\in H$. Thus we have $$\begin{align} (a,c)(b,d)&=(ab,cd) \\ &=(ba,dc) \\ &= (b,d)(a,c). \end{align}$$ Thus $G\oplus H$ is abelian.

Next we prove if $G\oplus H$ is abelian,then both $G$ and $H$ are abelian. Pick $x,y\in G\oplus H$. Let $(a,b)=x$. Let $(c,d)=y$. Thus we have $$\begin{align} (ab,cd)&=(a,c)(b,d) \\ &=(b,d)(a,c) \\ &=(ba,dc). \end{align}$$ Thus we have $ab=ba$ and $cd=dc$. Thus we know $G$ and $H$ are abelian.

Next, we provide a counterexample to show if both $G$ and $H$ are cyclic, then $G\oplus H$ is not necessarily cyclic. Let $G=Z_2$. Let $H=Z_2$. Thus we have $G\oplus H=\{(0,0),(0,1),(1,0),(1,1)\}$. Note how each element has order 2 and that the order of $G\oplus H$ is 4. Thus, because there does not exist any element in $G\oplus H$ that has order $|G\oplus H|$ we know that $G\oplus H$ is not cyclic.

Lastly, we prove if $G\oplus H$ is cyclic, then both $G$ and $H$ are cyclic. Let $(a,b)$ be the generator of $G\oplus H$. Pick $x\in G\oplus H$. Pick $c$ such that $x=(a,b)^c=(a^c,b^c)$. Let $(g_1,h_1)=x$. Thus we have $(g_1,h_1)=(a^c,b^c)$. Thus we know every element of $G$ can be written as $a^k$ for some $k\in\mathbb{Z}$. Therefore, $G$ is cyclic. Likewise, we know every element of $H$ can be written as $b^k$ for some $k\in\mathbb{Z}$. Thus $H$ is cyclic.

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