If $|G|=8$ Is Cyclic Then There Is $H\leq G$ Such That $|H|=4$ And $H$ Is Cyclic

Asked Mar 04 '20 at 10:41

Active Mar 04 '20 at 11:15

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Let there be $G$ a group such that $|G|=8$.

Prove: If $G$ is cyclic that it has a cyclic subgroup of order $4$

The options for the orders of a subgroup is $1,2,4,8$ so $4$ is a possibility.

Now $G=\{e,g,g^2,g^3,g^4,g^5,g^6,g^7\}$ has it is cyclic, taking $\langle \,g^2\mkern2mu\rangle=\{e,g^2,g^4,g^6\}$ is a cyclic subgroup of order $4$

I am asked, why is it obvious that the subgroup is cyclic? I know that every subgroup of a cyclic group is again cyclic, is there another reason here?

edited Mar 04 '20 at 11:15

Shaun

44,997

asked Mar 04 '20 at 10:41

newhere

3,115

2

If you know that every subgroup of a cyclic group is cyclic, then you don't need another reason. Of course, the subgroup $\langle g^2\rangle$ is cyclic also because it is generated by a single element, namely $g^2$. There is nothing much going on here. – Jyrki Lahtonen Mar 04 '20 at 10:43
Is it obvious that the subgroup is cyclic? Yes, if you show that $g^2$ has order $4$. But since you know the order of $g^n$ in a cyclic group, see this duplicate, it is obvious. – Dietrich Burde Mar 04 '20 at 10:44
@newhere You wrote $<g^2>$. What is it? – Michael Rozenberg Mar 04 '20 at 10:45
@MichaelRozenberg That means "generated by". So it is the cyclic subgroup generated by the element $g^2$ – Luke Mar 04 '20 at 10:46
@Luke My question was for the topic-starter. – Michael Rozenberg Mar 04 '20 at 10:47
@MichaelRozenberg Because it is only subgroup I could think of of order $4$ – newhere Mar 04 '20 at 10:49
1

newhere, the group $C_2\times C_2$ is not cyclic and also has order $4$. – Dietrich Burde Mar 04 '20 at 10:50

If $|G|=8$ Is Cyclic Then There Is $H\leq G$ Such That $|H|=4$ And $H$ Is Cyclic

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