coraletsmmh

2022-04-23

Product of two cyclic groups is cyclic iff their orders are co-'

Say you have two groups$G=\u27e8g\u27e9$ with order n and $H=\u27e8h\u27e9$ with order m. Then the product $G\times H$ is a cyclic group if and only if $gcd(n,m)=1$

Say you have two groups

Jonas Dickerson

Beginner2022-04-24Added 22 answers

Step 1

Note that $|G\times H|=\left|G\right|\left|H\right|=nm$; so $G\times H$ is cyclic if and only if there is an element of order nm in $G\times H$

In any group A, if $a,\text{}b\in A$ commute with one another, a has order k, and b has order $\ell $ then the order of ab will divide $\text{lcm}(k,\text{}\ell )$ (prove it).

Now take an element of $G\times H$, written as ${g}^{a},\text{}{h}^{b})$, where $G=\u27e8g\u27e9,\text{}H=\u27e8h\u27e9,\text{}0\le an,\text{}0\le bm$. Then $({g}^{a},{h}^{b})=({g}^{a},1)(1,{h}^{b})$

In this case, what is the order? Under what conditions can you get an element of order exactly nm, which is what you need

drenkttj9

Beginner2022-04-25Added 20 answers

Step 1

$B{\mathbf{Z}}_{m}\times {\mathbb{Z}}_{n}$ is noncyclic

$\iff \text{}B{\mathbf{Z}}_{m}\times B{\mathbf{Z}}_{n}$

$\iff \text{}\left\{rm\text{lcm}\right\}(m,n)mn$

$\iff \text{}!gcd(m,n)1$

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