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If an element $cinmathbb{U}_m$ has order $o$, then $mathbb{U}_m$ has elements of every order less than $o$

I’m trying to prove that, if some $cinmathbb{U}_m$ has order $o$ (i.e. $c^o = 1$), $mathbb{U}_m$ contains elements of every order less than $o$. It seems to me that perhaps this has something to do with $mathbb{U}_m$ being a cyclic group, however I’m not sure where to begin with this.

Edit: $mathbb{U}_m$ is the set of units in $mathbb{Z}_m$

Mathematics Asked by K_M on November 12, 2021

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