In a finite group the number of non identity elements which satisfy the given equation.$$ x^n =e$$. Where $e$ is identity. My efforts: If $n$ is prime then the number of elements are $multiple \ of \ (n-1) $. I generalised it by some examples. If n is not prime then what are the number of elements. Is it true that it will be a multiple of $\phi (n)$. If yes then how. Anyone can help.
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Does $n$ have any relation to the number of elements in the group? – Arthur Sep 26 '17 at 10:51
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@Arthur no n is just exponent i.e. belongs to natural number – Girish Kumar Chandora Sep 26 '17 at 10:52
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Well, any group of $4$ elements and $n = 4$ shows you that $\phi(n)$ is not involved, at least not as directly as you propose, since there are three solutions and $\phi(4) = 2$. In fact, if $n$ is the number of elements in the group, then $x^n = e$ is true for any element in the group, so there are always $n-1$ solutions. – Arthur Sep 26 '17 at 10:53
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@Arthur assumed that no of elements are larger than exponent. Example number of elements in a finite group with equation $x^4=e$ are multiple of 2 or 3or anything else? – Girish Kumar Chandora Sep 26 '17 at 10:57
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If $k$ is large then $S_k$ contains a lot of $4-$ cycles. More simply, just take a big product of cyclic groups of order $4$. – lulu Sep 26 '17 at 11:00
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@lulu lets modify the question the number of elements in a finite group with satisfy the equation $x^6=e$ will be? – Girish Kumar Chandora Sep 26 '17 at 11:02
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Nothing in what I wrote was specific to $4$. You can replace it with any exponent you like. – lulu Sep 26 '17 at 11:03
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@Arthur if $x^5=e$ then the number of elements are multiple of 4n. Similiarly what can we say about number of elements if $x^6=e$? – Girish Kumar Chandora Sep 26 '17 at 11:05
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@lulu if $x^5=e$ then the number of elements are multiple of 4n. Similiarly what can we say about number of elements if $x^6=e$? For $x^5=e$ this is the question of Joesph Gallian. I want to find the number of elements that satisfy $x^6=e$ – Girish Kumar Chandora Sep 26 '17 at 11:07
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The only result I am aware of along the lines you are asking is an old theorem of Frobenius which says that, assuming $n$ divides the order of a group, then the number of solutions to $g^n=e$ in the group is a multiple of $n$. See, for example, this – lulu Sep 26 '17 at 11:13
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@lulu thnks for sharing. But sorry. At present i only read the starting to chapters of Gallian book so i didnt get the stuff you shared. At present can u please tell me the number if elemnets which satisfy the given equation i.e example $x^6=e$ will be multiple of $2$? – Girish Kumar Chandora Sep 26 '17 at 11:18
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Yes, that number will be a multiple of $6$ (by the Theorem I cited). – lulu Sep 26 '17 at 11:20
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@lulu if multiple of 6 then it is failed for prime number. But for prime number i am definitely sure that it is multiple of (p-1). That is for example$ x^11=e$ the number of elemnts are multiple of 10. If $x^6=e$ the according to you number of elemnts are multiple of 6. Please can you prove it. – Girish Kumar Chandora Sep 26 '17 at 11:25
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What do you mean "failed for prime number"? In the cyclic group of order $p$ there are exactly $p$ solutions to $x^p=e$. in particular, in the cyclic group of order $11$, there are $11$ solutions to $x^{11}=e$. – lulu Sep 26 '17 at 11:29
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@lulu i appologies that i am not getting you because i am new to algebra . Can you please explain the number of elements which satisfy $x^6=e$ are multiple of 6 and what are they. For example $x^3=e$ has number of elements that are multiple of 2. And the elements are $x,x^2$. Can you list the elements for $x^6=e$ as i listed? – Girish Kumar Chandora Sep 26 '17 at 11:34
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I think we are talking at cross purposes. You keep writing about the equation $x^k=e$ but perhaps you mean to write about the number of elements of order exactly $k$. These are related concepts, obviously, but they are not equivalent. It is not true that the number of solutions to $x^3=e$ is a multiple of $2$. In the cyclic group of order $3$ there are $3$ solutions. – lulu Sep 26 '17 at 11:36
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@lulu please see this. Read "https://math.stackexchange.com/questions/845453/show-that-number-of-solutions-satisfying-x5-e-is-a-multiple-of-4" – Girish Kumar Chandora Sep 26 '17 at 11:55
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Exactly. That question specifies "non-identity". I am speaking of the total number of solutions, with no exclusions. For primes, that's just a matter of adding or subtracting $1$. For composites, you have to be clear what you are excluding. – lulu Sep 26 '17 at 12:12
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@lulu see my question i highlighted $non identity$ elements. Anyway can you please tell me the number of non identity elements which satisfy $x^4=e$. I generalised for prime exponents. Can we generalised for non prime exponents or not. I hope at this time you will get me. – Girish Kumar Chandora Sep 26 '17 at 12:15
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All we know is the Frobenius theorem. The number of solutions to $x^4=e$ is divisible by $4$. Call that number $4k$. $4k-1$ is not necessarily divisible by $3$. In the quaternion group of order $8$, there are $7$ solutions to $x^4=e$, excluding $x=e$. – lulu Sep 26 '17 at 12:20
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@lulu in your $7$ solutions is it include identity? If yes then the non identity solutions are 6 and which is mutiple of phi(6). If identity is not included then i am satisfied by you and the conversation closed here. – Girish Kumar Chandora Sep 26 '17 at 12:34
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It is not included. The group is non-cyclic (not even abelian) so every element in the group satisfies $x^4=e$. Excluding the identity that means there are exactly $7$ solutions. – lulu Sep 26 '17 at 12:44
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1@lulu thankyou for explaining and giving your precious time. Now i got it that my generalisation is wrong. – Girish Kumar Chandora Sep 26 '17 at 12:46