$\begingroup$Context
I am an undergrad student taking Abstract Algebra I, and this is a problem on a homework assignment.
Problem
Let $H$ be a normal subgroup of $G$, and let $a ∈ G$. If the coset $aH$ has order 3 in the factor group $G/H$, and $|H| = 10$, what are the possible orders of $a$ in $G$?
What I know
- Since $H$ is a normal subgroup, its left coset should equal its right coset.
- Since $H$ has 10 elements, the highest order of an element should be 10.
2 Answers
$\begingroup$Hint:
The order of $a^3$ is a divisor of the order of $H$ (Lagrange's theorem). On the other hand, if the order of $a$ is $d$, the order of $a^3$ is $\dfrac d{\gcd(d,3)}$. Can you deduce the possibilities for $d$?
$\endgroup$$\begingroup$By definition, the order of the coset $aH$ in the quotient group $G/H$ is the least positive integer $n$ such that $(aH)^n=H$, being $H$ the identity of $G/H$. But, by normality of $H$ in $G$, it is $(aH)^n=a^nH$ (induction on $n$), so $a^nH=H$ or, equivalently, $a^n\in H$. So, in your case, $a^3\in H$ and $a,a^2\notin H$. Now, $e=(a^3)^{|a^3|}=a^{3|a^3|}$, and hence $|a|$ divides $3|a^3|$. Since $|a^3|$ divides $|H|=10$, $|a^3|$ can be any among $1,2,5,10$, whence $3|a^3|$ can be any among $3,6,15,30$. To sum up, $|a|$ can be any among $3,5,6,10,15,30$, being $1$ and $2$ ruled out by the conditions $a,a^2\notin H$.
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