Group Theory – Number of Non-Isomorphic Non-Abelian Groups of Order 10

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Number of non-isomorphic non-abelian groups of order 10

Let $G$ be a group of order 10.Let $a\neq e \in G$ then it is not possible that all elements are of order 2 otherwise $G$ becomes abelian.

Let $a$ has order 5.then $H=\langle a\rangle$ then $H$ has order 5. Now $H$ will have two cosets say $bH,H$ Now $b^2\in H$ then $b^2$ is one of $e,a,a^2,a^3,a^4$ but $b^2\neq a^i$ for $i=1,2,3,4$ otherwise $o(b)=10$ contradiction

Again $bab^{-1}\in H$ as $H$ is normal also $bab^{-1}$ is one of $e,a,a^2,a^3,a^4$

Now I am getting possibilities of $bab^{-1}$ to be $a^2,a^3,a^4$

Does that mean there are 3 non-isomorphic non-abelian groups of order 10?

Best Answer

You should notice that when you conjugate $a$ with $b$ twice, you get back $a$. This gives you a constraint that allows you eliminate two of the alterenatives. Extended hints:

  • Given that $bab^{-1}=a^k$ for some $k$. Show that this implies that $ba^jb^{-1}=a^{jk}$ for all integers $j$.
  • Show that as a consequence of this $b^2ab^{-2}=a^{k^2}.$
  • Show that this leaves you with a single choice for $k$.
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