Prove that an element and its inverse have same order in any group.
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2Hint: $(g^{-1})^k = (g^k)^{-1}$ – Prahlad Vaidyanathan Dec 10 '13 at 11:08
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4@ user114830: You have to show what you tried first so people can gauge where you stand and help you accordingly. – wannadeleteacct Dec 10 '13 at 11:11
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the cycle group generated by g has $g^{-1}$ as an element ... – kjetil b halvorsen Dec 10 '13 at 11:15
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@kjetilbhalvorsen : I am sorry but i could not understand what you mean.... +1 at manasi – Dec 10 '13 at 11:25
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@YACP Do you really think mine was an answer warranting a downvote? If the definition of order is, as I like to do, the number of elements in the generated subgroup, then that's a full answer, but of course almost trivial; if the definition is based on what power gives the identity (as, unfortunately, is usually spelled out), then it's no more than a hint. – egreg Dec 10 '13 at 12:04
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2@arbautjc It's technically off-topic for the reasons explained below; the OP should add details about his/her work on the subject, in order to make the question more helpful for future readers. – egreg Dec 10 '13 at 17:05
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"Ok, but then many questions would be" Yes, and many are. "As long as it's mathematically sound" No (read the guidelines). – Did Dec 11 '13 at 12:36
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An element of a group has the same order as its inverse – Martin Sleziak Jul 11 '15 at 17:44
2 Answers
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Since $g^{-1}\in \langle g\rangle$, we have that $$\langle g^{-1}\rangle\subseteq\langle g\rangle$$ For the same reason $$\langle g\rangle\subseteq\langle g^{-1}\rangle$$
The order of an element $g$ is infinite if $\langle g\rangle$ is infinite, otherwise it's the same as $|\langle g\rangle|$ (number of elements).
egreg
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$$(g^{-1})^{o(g)}=(g^{o(g)})^{-1}=e^{-1}=e$$ proves that $o(g^{-1})\leq o(g)$.
$$g^{o(g^{-1})}=((g^{-1})^{-1})^{o(g^{-1})}=((g^{-1})^{o(g^{-1})})^{-1}=e^{-1}=e$$ proves that $o(g)\leq o(g^{-1})$
Danae Kissinger
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