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Let $G_1,G_2$ be two finite cyclic groups with orders $\text{ord}(G_1)=n_1, \text{ord}(G_2)=n_2$ and $G:=G_1\times G_2$. Why is $\text{ord}(G)=n_1\cdot n_2$ if $G$ is cyclic?

I know that there are $a_1,a_2 \in G_i$ with $G=\langle a_1,a_2\rangle $, but which theorem says that $\text{ord}(a_1,a_2)=\text{ord}(a_1)\cdot \text{ord}(a_2)=n_1\cdot n_2$?

Shaun
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KingDingeling
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  • Strictly speaking, $G$ is the set product of two finite groups, and the number of elements in $G$ is equal to $|G_1|\cdot|G_2| = n_1\cdot n_2$. – Good Morning Captain Jan 12 '20 at 15:00
  • The order of $G$ is the product of the orders of $G_1$ and $G_2$ without any other assumptions. – the_fox Jan 12 '20 at 15:00
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    You may be confusing order of a group with order of an element. The order of $(a_1,a_2)$ is the least common multiple of the orders of $a_1$ and $a_2$, which in general is not the product of the orders, but that just says the product of two cyclic groups is in general not a cyclic group. – Gerry Myerson Jan 12 '20 at 15:08

1 Answers1

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$|G_1 \times G_2| = |G_1| \times |G_2| = n_1 n_2$, which follows from the formula for the cardinality of the cartesian product of sets. Also, $G$ is only cyclic if $\gcd(n_1,n_2) = 1$, see here

Ryan Shesler
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