Suppose that the $o(gH) = n \in \frac{G}{H}$ where $\frac{G}{H}$ is the quotient group. Show that if the $o(g) = m \in G$ then $n$ divides $m$
I want to use a version of Lagrange's theorem here, but I feel like that might not apply. Because I know the number of cosets is equal to the index, but I don't know the order of the entire group.
Should I more use the definition of order and start the proof with something like $$(gH)^n = g^nH$$ but we know that $o(g) =m$ so $m$ must be the smallest element so that $g^m =e$ so we then know that $n$ and $m$ are divisors, but that $n$ must be greater than $m$
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$\begingroup$OK, since I seem to have made a muddled mess in the comments:
Suppose $o(g) = m$. This means that $g^m = e$ ($m$ isn't the only positive integer with this property, but it is the least such).
Now $(gH)^m = g^mH = eH = H$, which is the identity of $G/H$.
It follows that $o(gH) \leq m$. Here, $m$ is one positive integer with $(gH)^m = H$, but there may be smaller ones.
Since $m \geq n = o(gH)$, we can write:
$m = qn + r$ for an integer $0 \leq r < n$.
Now $H = (gH)^m = (gH)^{qn+r} = (gH)^{qn}(gH)^r = ((gH)^n)^q(gH)^r = H^q(gH)^r = (gH)^r$.
If $r \neq 0$, this contradicts $n = o(gH)$ (since $r$ would be a smaller positive number than $n$ with $(gH)^r = H$), so we conclude $r = 0 \implies m = qn$, that is, $n\mid m$.
Here is an example where $n < m$:
Consider the group $G = \{e,a,a^2,a^3,a^5\}$ which is cyclic of order $6$. We can use $g = a$, which has order $6$.
We have the (normal) subgroup $H = \{e,a^3\}$. Explicitly, here are the cosets of $G/H$:
$H = \{e,a^3\}$ (this is also $a^3H$).
$aH = \{a,a^4\}$ (this is also $a^4H$)
$a^2H = \{a^2,a^5\}$ (this is also $a^5H$). so $G/H = \{H,aH,a^2H\}$.
So now let's look at the order of $aH$.
Since $aH \neq H$, it is clear $aH$ does not have order $1$.
$(aH)^2 = a^2H \neq H$, so $aH$ does not have order $2$.
$(aH)^3 = a^3H = H$ (see above), so $o(aH) = 3$, even though $o(a) = 6$.
Now $(aH)^6 = a^6H = eH = H$, as in the argument above, and indeed, we see $3\mid 6$.
The reason $o(gH) \leq o(g)$ is that we do not need $g^n = e$ (as in $G$), but only $g^n \in H$.
$\endgroup$ 1 $\begingroup$Hint: If $G$ is any group, $g\in G$ has order $n$ and $g^m=e$, then $n\mid m$.
If not, then $$m=qn+r$$for some $1\le r\lt n$, so $$g^r=g^{m-qn}=g^m(g^{qn})^{-1}=e$$which contradicts the fact that the order of $g$ is $n$.
Can you apply this here?
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