If we have an element $a$ of order $p-1$ and an element $b$ of order $p^{e-1}$ in $\left( \mathbb{Z} \backslash (p-1)p^{e-1} \mathbb{Z} \right)^{\times}$. From which theorem it follows that the order of $a*b$ is $(p-1)p^{e-1}$?
Thanks in advance.
Order of $a*b$ is $(p-1)p^{e-1}$ if order of $a$ is $(p-1)$ and order of $b$ is $p^{e-1}$.
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group-theory
number-theory
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0In $a$ and $b$ are elements of (finite) orders $m$ and $n$ in any commutative group, then the order of $ab$ is $\text{lcm}(m,n)$. This fact is rather easy to prove. I'd not call it "theorem". – 2017-02-16
1 Answers
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It follows from this theorem about the order of a product $ab$ in an abelian group. If $ord(a)=m$, $ord(b)=m$, then $ord(ab)=lcm(m,n)$. Since $gcd(p-1,p^{e-1})=1$, the lcm is just the product.
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0Okay,, thanks!! – 2017-02-16