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I want to show that $ D_{2mn} $ is not isomorphic to $ D_{n} \times D_{m} $.

First, if both $m,n$ are even, then an element of maximum order in $D_{2mn}$ has order of $2mn$, but in the product, we can see that $s^ir^j$ (in both $D_n, D_m$) has order of at most $mn$.

However, I'm having difficulty finding the number of elemnts of order $mn$ for $m,n$ odd: in $D_{2mn}$, $sr^j$ is never of order $mn$; but what of $r^{2i}$? I know $r^2$ is of that order, but I'm not sure about all the others; and I don't know how to find the number of elements in the cartesian product of order $mn$.

Could I receive some ideas? Is there a different way to prove it? Thank you.

FNB
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2 Answers2

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It suffices to show that the product $D_n \times D_m$ contains no element of order $2nm$. Take any element $(x,y)$, where $x \in D_n$ and $y \in D_m$. It is a basic fact that the order of $(x,y)$ is the lowest common multiple of the orders of $x$ and $y$.

Now the order of $x$ either is $2$ or divides $n$, and that of $y$ either is $2$ or divides $m$. You can check that for each of the four cases here, the order of $(x,y)$ can never be greater than $nm$. In particular, it cannot have order $2nm$.

In your question, you also ask how to count the number of elements of order $nm$. This is not a very useful method to solve the problem, but I will write a short discussion on it anyway. From the above, we deduce that such elements arise in $D_n \times D_m$ if and only if $n$ and $m$ are coprime. In that case, the number of such elements is precisely the number of elements in $D_n$ of order $n$, times the number of elements in $D_m$ of order $m$. The answer can be given in terms of the Euler totient function as $\varphi(n)\varphi(m)$. (In fact, this is equal to $\varphi(nm)$. See What's the proof that the Euler totient function is multiplicative?)

(For more details on the last paragraph, see How to find a generator of a cyclic group?)

George Wood
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Assume that $D_{2mn}\cong D_m \times D_n$, with $m,n\ge 3$. Then $D_n$ and $D_m$ are normal subgroups of $D_{2mn}$. However, the only nonabelian normal subgroups are $D_{2mn}$ itself and $D_{mn}$, see this post. This is a contradiction, considering the cardinalities.

What is true is, that $D_{2n}\cong D_n\times C_2$, for odd $n\ge 3$.

Dietrich Burde
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