I was reading in MSE that, up to isomorphism, there are 2 groups of order 45. How do we know that? Is there any way of calculating how many groups of order 10,15 etc. exist up to isomorphism?


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  • $\begingroup$ Do you know Sylow theorems in group theory? $\endgroup$ – DonAntonio Jun 2 '16 at 9:54
  • $\begingroup$ Yes. I know three theorems of sylow and that means only one sylow 3 and sylow 5 subgroup. $\endgroup$ – low iq Jun 2 '16 at 10:40
  • $\begingroup$ But I am asking a general way.would sylow theorem be enough for all orders? $\endgroup$ – low iq Jun 2 '16 at 10:40
  • $\begingroup$ Or "special orders " only? $\endgroup$ – low iq Jun 2 '16 at 10:42
  • $\begingroup$ It's easy to count the abelian groups of this order by the classification of finite abelian groups. From there, it suffices to show that there are no non-abelian groups of this order. $\endgroup$ – Omnomnomnom Jun 2 '16 at 11:09

As you say, by Sylow theorems a group $\;G\;$ of order $\;45\;$ has one unique subgroup $\;P\;$ or order $\;3^2=9\;$ and one unique subgroup $\;Q\;$ of order $\;5\;$ , which means $\;P,\,Q\lhd G\;$ . Also, both $\;P,\,Q\;$ are abelian and so is their product, which generates their direct product since $\;P\cap Q=\{1\}\;$ , and from all this it follows that

$$G=PQ=P\times Q$$

and $\;G\;$ is thus always abelian. Since there are two groups up to isomorphism of order $\;p^2\;$ for any prime $\;p\;$ , we get two unique (up to isomorphism) different groups of order $\;45\;$ (both abelian, again):

$$G_1=C_9\times C_5\;\;,\;\;\;G_2=C_3\times C_3\times C_5\cong C_3\times C_{15}$$

The above is a particular case of the general: if $\;p<q\;$ are two primes such that $\;p\,\nmid\,q-1\;$ , then there are two unique groups of prder $\;p^2\cdot q\;$ up to isomorphism

  • $\begingroup$ How do you know that p intersection Q ={1} $\endgroup$ – low iq Jun 2 '16 at 14:22
  • $\begingroup$ Can you please tell me where would I find more on this? in easy way ! $\endgroup$ – low iq Jun 2 '16 at 14:23
  • $\begingroup$ @lowiq We have that $\;P\cap Q=1\;$ because any element in the intersection must divide both $\;p\;$ and $\;q\;$ , and these two are different prime numbers... About more exercises: any book in group theory surely will contain some exercises of this, and in the internet there must be thousands of sites... $\endgroup$ – DonAntonio Jun 2 '16 at 16:18

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