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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.

  1. The order of $a$ in $G$ is a multiple of the order of $Na$ in $G/N$.
  2. 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.

Solution

1.

Let $a \in G$ and $Na \in G/N$, and let $n = |a|$. Let $f: G \rightarrow G/N$ defined by $f(g) = Ng$ be a function (we know that $f$ is a function because if $x = y$ for $x,y \in G$, then clearly $x = y \in Ny$ showing that $Nx = Ny$). Let $b,c \in G$. We compute $$ \begin{align} f(bc) &= N(bc) \\ &= (Nb)(Nc) \\ &= f(b) f(c), \end{align} $$ showing that $f$ is also a homomorphism between $G$ and $G/N$. Therefore, we know that $e_H = f(e_G) = f(a^n) = f(a)^n = Na^n$. Thus we have that $e_H = Na^n$, showing that $n$ is a multiple of $k$.

2.

Given that $n$ is a multiple of $k$ we may write $n = m k$ for some $m \in \mathbb{Z}$. We claim that $b = a^m \in G$ has order $k$. We compute $$ \begin{align} |b| &= |a^m| \\ &= \frac{n}{\gcd(m, n)} \\ &= \frac{mk}{\gcd(m, mk)} \\ &= \frac{mk}{m} \\ &= k. \end{align} $$ $\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \blacksquare$

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