Finding the left cosets for cyclic group $G=\langle a\rangle>$ of order $12$ and the cyclic subgroup $H=\langle a^4\rangle$

Question

I am trying to check my work mostly and if not then could I get some hints,

First I am asked to find the left cosets of $H$ in $G$. So I need to find the elements in $H$. To do this I use a theorem in my textbook but I do not know if I really need the theorem to do this:

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$Theorem$:

Let $G$ be a finite group and let $x\in G$. Then $o(x)$ divides $|G|$. Consequently, $x^{|G|}=e$ for every $x\in G$.

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From this, take $a\in G$ then $o(a)=|G|=12$. By this we then have $a^{12}=e$ identity. This leads me to see that $H=<a^4>=${$a^4,a^8,e$} so our left cosets should be:

$aH$={$a^5,a^9,a$}

$a^2H$={$a^6,a^{10},a^2$}

$a^3H=${$a^7,a^{11},a^3$}

$a^4H=${$a^8,e,a^4$}$=H$

is this correct?

Answer 1

You are correct that $H=${$a^4, a^8, e$}.

If $G$ is a group and $H$ is a subgroup of $G$,

then the left cosets of $H$ in $G$ are of the form $gH$, i.e., {$gh|h \in H$}.

The elements $g\in G$ in this example are $e, a, a^2, a^3, a^4, a^5, ..., $ and $a^{11}.$

So the left cosets in this example are $eH, aH, a^2H, a^3H, a^4H, a^5H, ..., $ and $a^{11}H,$

but $a^4H$ is the same as $H$, $a^5H$ is the same as $aH$, ...