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Let $G$ a group where $|G|=15$, let's also suppose that $G$ is not abelian, we also have an action from $G$ on itself by conjugation, I've proved that $G$ has exactly one orbit of 5 elements, three orbits $O_1$,$O_2$,$O_3$ of three elements and the orbit $\{e\}$, I have two questions please :

1) How do I prove that elements of the same orbit have the same order?

2) How do I find the order of each element in each Orbit?

If you'd help me it would be great, thank you very much.

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(1) elements in the same orbit are, by definition of your action, conjugates in $G$. So they have the same order.

(2) You eliminate the possible orders. Take the orbit $O$ of 5 elements, for example, and suppose $g\in O$. Then the order of $g$ cannot be 15 (otherwise the group is abelian) or 1 (that is the identity), so must be either 3 or 5. If it has order 5, then the orbits of $g^2,g^3,g^4$ also have the same lengths. So $g,g^2,g^3,g^4\in O$. Now what is the other element in $O$? and you get a contradiction. Similar for the other cases.

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  • $\begingroup$ thank you this has been a great help $\endgroup$ – Blueberry Jun 7 at 10:58
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This was getting too long and technical for a comment.

One observation you might make is that in any group there are an even number of elements of each order other than $1$ or $2$ - you can pair $x$ with $x^{-1}$ which has the same order. (This incidentally implies that the number of elements of order $2$ in any finite group is odd)

There are two cases for a non-identity element $x$ of odd order

Case A: $x$ and $x^{-1}$ are in the same conjugacy class

The class contains the inverse of all its elements and therefore has even size. So, if the class contains $y=axa^{-1}$ then it contains $y^{-1}=ax^{-1}a^{-1}$.

Case B: $x$ and $x^{-1}$ are in different conjugacy classes

These classes have the same size because we can pair $axa^{-1}$ with its inverse $ax^{-1}a^{-1}$ in the other class. Two conjugates of $x$ are equal, iff their pairs are equal. We therefore have two conjugacy classes which are the same size.

If the group has odd order, Case A cannot happen (it requires a conjugacy class of even size), so we must be able to pair all the classes except the class of the identity. This requires an even number of classes of each size (other than size $1$ where we have an odd number).

Also note that in a group of odd order $x$ and $x^{-1}$ are therefore in different conjugacy classes (except for the identity).

You don't need all this for your question on groups of order $15$, but it might help if you are tackling other similar problems.

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  • $\begingroup$ thank you for raising so many problems that i might have seen without understanding them $\endgroup$ – Blueberry Jun 9 at 14:17

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