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If I have a group $G$ and I take two subgroups $\langle g\rangle$ and $\langle g^{-1}\rangle $. Are their orders equal?

So far, I am just trying to find counterexamples. I worked with dihedral groups, integer modulos, etc. but all of them seemed to work well and the orders seem to be same.

How do I prove this for general case if it is true?

Arnaud D.
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Four Seasons
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    "...and..."?${}{}{}{}{}$ You ask about orders in the title, but then you talk about equal subgroups...Which one is it? – DonAntonio Feb 12 '18 at 11:21
  • I meant the order of is equal to order of <g^-1> – Four Seasons Feb 12 '18 at 11:22
  • The order of is the order of g, by definition. – Arnaud Mortier Feb 12 '18 at 11:23
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    Then why do you ask about subgroups? And yes: the order of $;g;$ equals that of $;g^{-1};$ . Try to prove it yourself, it isn't hard. – DonAntonio Feb 12 '18 at 11:25
  • @DonAntonio I think the op does not aware to the notion of little oh and big oh, for him $o$ is for order, so he asks whether the order of $g$ is equal tot he order of $g^{-1}$ – Yanko Feb 12 '18 at 11:27
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    @ArnaudMortier No, it isn't. Those are, at least by definition, different concepts, although it's rather easy to show that the two numbers coincide. The order of $g$ is the smallest positive power of $g$ which gives the identity element. The order of $\langle g\rangle$ is the cardinality of the set ${\ldots,g^{-1}, e, g, g^2,\ldots}$. – Arthur Feb 12 '18 at 11:29
  • @yanko Tha's exactly what I addressed and answered. I didn't mean at all little "o"...in fact, I couldn't tell what is little "o" within general group theory... – DonAntonio Feb 12 '18 at 11:31
  • @Arthur you're right. By definition here meant "by definition and easy arguments". – Arnaud Mortier Feb 12 '18 at 11:31

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