[Math] Show that for a $∈ G, Na ∈ G/N$, order(Na) is a factor of order(a).

abstract-algebragroup-theory

Let $(G, ·$) be a group and $N$ be a normal subgroup and $G/N$ be
the corresponding quotient. Show that for a $∈ G, Na ∈ G/N$

order(Na) is a factor of order(a).

I am really confused about how to proceed with this. I understand that it is asking me to prove that the order of an element in the quotient group divides the order of an element in a group that it is the quotient group of, but I am unsure of how to proceed.

Here is what I have so far:

Assume that $ord(a) = m $

$(Na)^m = N(a^m) = N(e) = Ne = N$

But i am not really sure where I can take this with the work above. Any help would be appreciated!

Best Answer

The proof is essentially the same as the proof that if $\psi:G\rightarrow H$ is a homomorphism, then for all $g\in G$, $o(\psi(g))$ divides $o(g)$ (provided $o(g)$ is finite).

Suppose that $o(g)=n$ and $o(\psi(g))=m$. Then, observe that $\psi(g)^n=\psi(g^n)=\psi(e)=e$. Since $o(\psi(g))=m$, it follows that $\psi(g)^m=e$. Let $d=\gcd(m,n)$, then there are integers $a$ and $b$ so that $d=am+nb$. Then $\psi(g)^d=\left(\psi(g)^m\right)^a\left(\psi(g)^n\right)^b=e^ae^b=e.$ Therefore, $o(\psi(g))\leq d$, but since $o(\psi(g))=m$, $m\leq d$. However, since $d=\gcd(m,n)$, $d\leq m$, so $d=m$. Since $\gcd(m,n)=m$, it follows that $m$ divides $n$.

Now, use the fact that $q:G\rightarrow G/N$ is a homomorphism and you're done. Alternatively, take the $\gcd$ part out of the argument and apply it directly in this case.

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