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I want to show that two cyclic subgroups are not equal. That the only element common to them is the identity element.

Is it enough to show that (1) their intersection is just the identity element OR (2) need to show that the generator of one cyclic subgroup is not contained on the other and vice versa?

I am doing approach (2). It would be simplier if (1) is enough. Thank you.

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  • $\begingroup$ What the heck? Yeah, approach #1. $\endgroup$ – Randall Jan 3 '18 at 5:29
  • $\begingroup$ You want to show that two groups are not equal and their intersection is trivial. Then yes, method $(1)$ works, provided of course the groups are not themselves the trivial groups. $\endgroup$ – Dave Jan 3 '18 at 5:30
  • $\begingroup$ Sorry @Randall. I'm really confused. Thank you. :) $\endgroup$ – Orior Jan 3 '18 at 5:31
  • $\begingroup$ Thank you very much @Dave. :) $\endgroup$ – Orior Jan 3 '18 at 5:31
  • $\begingroup$ No sorry, just keep it simple. Subgroups are subsets, so to show two sets are different you... $\endgroup$ – Randall Jan 3 '18 at 5:33
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You are actually asking two non-equivalent questions:

Question 1. Are the two cyclic subgroups different?

Question 2. Is the intersection of the cyclic subgroups trivial?

A positive answer to Q2 would give a positive answer to Q1 provided that the two subgroups are not trivial. But a positive answer to Q1 does not necessarily give a positive answer to Q2. Now Approach (1) is appropriate for Q2. For Q1, you may modify Approach (2) by deleting the "and vice versa" part.

Finally, be aware that sometimes the term "are equal" is used in place of "are isomorphic", which would lead to a third question, for which none of your approaches would suffice.

Question 3. Are the two cyclic subgroups isomorphic?

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  • $\begingroup$ Wow! Thanks much. $\endgroup$ – Orior Jan 3 '18 at 6:27
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    $\begingroup$ You're welcome. $\endgroup$ – J.-E. Pin Jan 3 '18 at 6:31

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