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Problem 93 (If A Factor Group $G/N$ Has An Element Of Order $k$ Then So Does $G$)
Suppose that $N$ is a normal subgroup of a finite group $G$ and that $Na$ has order $k$ as an element of $G/N$. Prove the following two facts.
- The order of $a$ in $G$ is a multiple of the order of $Na$ in $G/N$.
- There exists an element $b\in G$ that has order $k=|Na|$.
In other words, if a factor group has an element of order $k$, the the original group before factoring must have an element of order $k$ as well.
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