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I am willing to show that every group of order 15 is cyclic, using class equation.

Let $G$ be a group of order 15. If $G$ is abelian, then $G=Z(G)$ and so for each $a\in G=Z(G)$ we have $cl(a)=\{a\}$. Hence only possible class equation is $$15=1+1+1+\cdots+1+1(15~\text{times})$$ In this case, $G$ is isomorphic to either $\mathbb{Z}_{15}$ or the external direct product $\mathbb{Z}_3\times \mathbb{Z}_5$. But since $\mathbb{Z}_{15}\simeq \mathbb{Z}_3\times \mathbb{Z}_5$, it follows that $G$ is cyclic.

We now show if $G$ is non-abelian then contradiction will appear.

When $G$ is non-abelian, then $G\neq Z(G)$. Here $|Z(G)|\in \{1,3,5\}$. But if $|Z(G)|=3$ then $|G/Z(G)|=5$ a prime, so $G/Z(G)$ is cyclic and hence $G$ becomes Abelian, contradiction.

Similalry $|Z(G)|\neq 5$ as well and so only possibility is $|Z(G)|=1$. Then the class equation reads $$15=|G|=|Z(G)|+\sum |cl(a)|=1+\sum |cl(a)|$$ where the sum is taken over the orders of all non-singleton conjugacy classes $cl(a)$ in $G$.

Let there be $x_3$ and $x_5$ number of conjugacy classes of order 3 and 5 respectively in $G$. Then we must have $$15=1+3x_3+5x_5\Rightarrow 14=3x_3+5x_5$$ which is satisfied by $x_3=3, x_5=1$ so that ultimately the class equation becomes $$15=1+(3+3+3)+5$$

I do not know how to bring contradiction here. Any help?

Thanks in well advance.

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    $\begingroup$ Elements of order $3$ come in pairs and elements of order $5$ in fours (consider the cyclic subgroups they generate). You can't therefore have an odd number of elements of either order. $\endgroup$ – Mark Bennet Aug 28 '20 at 18:01
  • $\begingroup$ @MarkBennet Pardon me, but $3+3+3$ means they are the cardinality of conjugate classes. Why should we consider them as order of some elements ? $\endgroup$ – KON3 Aug 28 '20 at 18:49
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    $\begingroup$ You don't have to - but you can't have nine elements of order $3$ or of order $5$, because the number of elements of each of these orders has to be even. $\endgroup$ – Mark Bennet Aug 28 '20 at 19:10
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    $\begingroup$ @Anjan3 I think what he is saying is that suppose $a$ is an element of order $3$ and let $A=\langle a \rangle$. Then $A$ has exactly two elements of order $3$ . If $b\notin A$ and $o(b)=3$ then we get additional two elements of order $3$ , namely $b$ and $b^2$ . Thus in total there are even no. of elements of order $3$. Similar arguments for elements of order $5$ . Take a conjugacy class having $3$ elements, $C$ (say). Now elements in the same conjugacy class has same order. So each elements in this conjugacy class $C$ has order $3$. $\endgroup$ – user-492177 Aug 28 '20 at 19:50
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    $\begingroup$ @Anjan3 No no. That was a typo error . I meant if $b\notin A$ and $o(b)=3$ and so on.... See my edit $\endgroup$ – user-492177 Aug 28 '20 at 20:06
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All the elements in a single conjugacy class have the same order. You have to have an even number of elements of order $3$ and the number of elements of order $5$ has to be divisible by $4$. We already have the identity element.

Look at the elements of order $5$ - there must be $4, 8, 12$ of these. This means $10, 6, 2$ elements of order $3$ (given none of order $15$)

The only possibility from what you have is $3+5=8$ elements of order $5$ and $3+3=6$ elements of order $3$.

There are various ways of working from here - but with $8$ elements of order $5$ there would have to be two subgroups of order $5$ and conjugation (consider the larger class) would have to take at least one generator of one to a generator of the other, and you should be able to see that this can't split the eight elements as $3+5$ for a contradiction.

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  • $\begingroup$ I understood your answer. Please clear my confusion on this. If we choose a conjugacy class of order 3, how can we say it will have elements of order 3? Is there any results related to that ? If so, please give me the link so that I can study that. $\endgroup$ – KON3 Aug 29 '20 at 13:49
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    $\begingroup$ @Anjan3 Conjugacy classes contain elements and it is the elements which have orders. Conjugate elements have the same order, so each conjugacy class is a set of elements having a common order. You can't tell directly from the size of a class the order of the elements it contains. Here, given the size of the classes you have found, you can't have a single class of elements of order $5$ (you need the number of elements to be a multiple of $4$) so you need to make $8$ elements from two classes. The remaining classes must contain elements of order $3$. $\endgroup$ – Mark Bennet Aug 29 '20 at 13:55
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    $\begingroup$ @Anjan3 In an abelian group conjugacy classes all have size $1$ regardless of the order of the elements. If you want to see some examples you might like to think about the conjugacy classes of the dihedral, symmetric and alternating groups of small order since these have relatively simple structures, and that will give you some reference points. $\endgroup$ – Mark Bennet Aug 29 '20 at 13:57
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Why $15 =1 + 3 + 3 + 3 + 5$ cannot be the class equation of group of order 15.

We know that number of elements of order $5$ can be $4,8,12$,..

And

Number of elements of order $3$ can be $2,4,6$,..

So the only way, the given class equation would make sense is by having $8$ elements of order $5$ which fit in class of sizes ($3,5$) and $6$ elements of order 3 which fit in class of sizes ($3,3$) ..

BUT ,if that's the case , since there are $8$ elements of order $5$, there are $2$ subgroups of order $5$ (H and K ) .. then HK order will be $25$, which is not possible (because HK is subset of G, G contains $15$ elements)

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  • $\begingroup$ nice arguments indeed $\endgroup$ – KON3 Mar 7 at 13:28

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