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As the title says I have: $G$ cyclic group of order 24, and $H= \langle x^6 \rangle$ , we also know that $x$ is a generator of $G$ so, $\langle x \rangle = G$

i have to find the order of each element of the quotient group $G/H$

and say how many generators have $G/H$

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    $\begingroup$ I'm the \langle \rangle fairy, here to let you know that $\langle, \rangle$ plays nicer with TeX than <, > does :) $\endgroup$ – Patrick Stevens Dec 4 '15 at 8:12
  • $\begingroup$ Is $H$ generated by $x^6$ for some particular $x$ you forgot to define or $H$ is generated by all elements of $G$ that are of the form $x^6$? $\endgroup$ – Clément Guérin Dec 4 '15 at 8:13
  • $\begingroup$ I think she forgot to mention that $x$ is a generator of $G.$ $\endgroup$ – Justpassingby Dec 4 '15 at 8:13
  • $\begingroup$ i think that $G=\{e,x,x^2,x^3,...,x^6,...,x^(23) \}$ since $G$ is a cyclic group. $\endgroup$ – Zigisfredo Dec 4 '15 at 8:15
  • $\begingroup$ @Zigisfredo it's not automatic, you need to mention that. If $x$ is an arbitrary element of $G$ then its order could be less than $24.$ $\endgroup$ – Justpassingby Dec 4 '15 at 8:17
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(answer adapted to some of the comments)

The order of $H$ is the order of its generator $x^6,$ i.e., 4. Therefore the order of $G/H$ is 6: there are 6 different cosets of the form $x^iH.$

We can choose the first powers of $x$ as representatives of these cosets:

$$G/H=\left\{H,xH,x^2H,x^3H,x^4H,x^5H\right\}$$

The order of an individual coset $x^iH$ is the smallest nonzero natural number $n$ such that $i.n$ is a multiple of 6.

Now at least one of the elements of $G/H$ turns out to have its order equal to the order of $G/H$ itself, implying that $G/H$ is cyclic.

A cyclic group is generated by an individual element $g$ iff the order of that element is the order of the group.

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  • $\begingroup$ i think that i found the order of $H$, since $H=\langle x^6 \rangle=\{e,x^6,x^12,x^18\}$ so the order of $H=4$ therefore the order of $G/H$ is 4. But after that i'm confused about how to get for each element the order. $\endgroup$ – Zigisfredo Dec 4 '15 at 8:24
  • $\begingroup$ No, the order of $G/H$ is determined by Lagrange's theorem and it's different from 4. $\endgroup$ – Justpassingby Dec 4 '15 at 8:25
  • $\begingroup$ oh, so i have, $o(G/H)=o(G)/o(H)=24/4=6$. $\endgroup$ – Zigisfredo Dec 4 '15 at 8:28
  • $\begingroup$ Yep, you're on your way! Now propose representatives of the 6 cosets of $H.$ $\endgroup$ – Justpassingby Dec 4 '15 at 8:29
  • $\begingroup$ Should i take $x^p$ with $p$ a prime number so i could get the order easier? $\endgroup$ – Zigisfredo Dec 4 '15 at 8:41

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