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Chain rule proof doubt

Given that $C_n$ is a cyclic group of order $n$, what conditions must integers $n$ and $m$ satisfy such that the group $C_n \times C_m$ is isomorphic to C$_{mn}$?

So I attempted to investigate a few lower-order cyclic groups to find a pattern. In particular, I looked at $C_2 \times C_3$ and compared it to $C_6$. Here are a couple of diagrams:

$C_2 \times C_3$ – http://escarbille.free.fr/group/?g=6_2b

- Let G be an abelian group, and let a∈G. For n≥1,let G := {x∈G:x^n =a}. Show that G is either empty or equal to αG := {αg : g ∈ G}…
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$C_6$ – http://escarbille.free.fr/group/?g=6_2a

They Cayley tables look pretty different (identity elements are all over the place for $C_2 \times C_3$) so I concluded that they are not isomorphic.

I did the same for $C_2 \times C_4$ and $C_8$:

$C_2 \times C_4$ – http://escarbille.free.fr/group/?g=8_2

$C_8$ – http://escarbille.free.fr/group/?g=8_1

Again, I arrived at the conclusion that the 2 groups are not isomorphic due to their Cayley table structure.

It seems I’m stuck now. Are my conclusions wrong for the examples I examined? If so, how do I find the conditions for the general case of comparing $C_m \times C_n$ with $C_{mn}$?

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The correct condition is that $m$ and $n$ are coprime, that is, have no common factors. In particular, this means that $C_2 \times C_3 \cong C_6$. (Sometimes it is hard to compare two multiplication tables by inspection, though it can help identifying two isomorphic groups from their tables by reordering the elements.)

**Hint** Given two groups $G, H$ and elements $a \in G, b \in H$, the order of element $(a, b)$ is the smallest number, $\text{lcm}(k, l)$ divisible by both the order of $l$ of $a$ and the order $k$ of $b$.

In particular, suppose $G$ and $H$ are cyclic, say $G = C_n$, $H = C_m$. Then, if $a$ and $b$ are generators of $C_n$, $C_m$, respectively, by definition they have respective orders $n, m$ and hence $(a, b) \in C_n \times C_m$ has order $\text{lcm}(m, n)$.

So, if $m, n$ are coprime, this order is $\text{lcm}(m, n) = mn$.

Conversely, if $m, n$ are not coprime, then since any $a \in C_n$ has order $l$ dividing $n$ and any $b \in C_m$ has order $k$ dividing $m$, we have that the generic element $(a, b) \in C_n \times C_m$ has order $\text{lcm}(k, l) \leq \text{lcm}(m, n) < mn$. In particular, no element has order $mn$, so $C_m \times C_n$ is not cyclic.

For example, the elements $1 \in C_2$ and $1 \in C_3$ generate their respective groups, and the powers of $(1, 1) \in C_2 \times C_3$ are

$$(1, 1), (0, 2), (1, 0), (0, 1), (1, 2), (0, 0),$$

so $(1, 1)$ generates all of $C_2 \times C_3$, which is hence isomorphic to $C_6$.

Proof for $m,n$ relatively prime:

Denote $C_n$ by $C_n = \Bbb Z_n = \{0,1,\dots,n-1\}$, where the group operation is taken to be addition (modulo $n$).

We then have the isomorphism

$$

\phi: \Bbb Z_m \times \Bbb Z_n \to \Bbb Z_{mn}\\

\phi(a,b) = na + mb

$$

I will leave it to you to verify that $\phi$ satisfies the defining properties of an isomorphism.

$C_2\times C_3$ is actually isomorphic to $C_6$.

In fact, you can map $(0,0)\mapsto 0$, $(1,0)\mapsto 3$, $(0,1)\mapsto 2$, $(1,1)\mapsto 5$, $(0,2)\mapsto 4$, $(1,2)\mapsto 5$ and create an isomorphism between the two groups.

Also, saying “The tables look different, so they are not isomorphic” is not a proper mathematical argument.

My advice is that you look at how the particular isomorphicm I wrote was created. Then, try to create a similar one on $C_4\times C_5$, for example (it is possible because $C_4\times C_5$ is also isomorphic to $C_{20}$).

Then, try to create the isomorphism between $C_4\times C_2$ and $C_8$ and see why that won’t work. Then you may get an idea what property the pair $m,n$ must have in order for $C_m\times C_n$ to be isomorphic to $C_{mn}$

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