Number of non-isomorphic non-abelian groups of order 10 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?
 A: 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 $k$.

A: The number of non-isomorphic subgroups of order 10 can be found using semidirect products:
Let $G$ be a non-abelian group of order 10 and let $H=\langle a \rangle$ be its unique Sylow 5-subgroup (that is $|H|=5$), which is normal as you claimed. Let $K=\langle x \rangle$ be a Sylow 2-subgroup of $G$ (that is $|K|=2$). Since


*

*$G=HK$

*$H \trianglelefteq G$

*$H \cap K=1$


we have that $G \cong H \rtimes_{\varphi} K$ is a semi-direct product for some homomorphism $\varphi:K \to \text{Aut}(H)$. This homomorphism is non-trivial, for otherwise $G$ would have been abelian, and also, it is determined completely by the image $\varphi(x)$. We can list all non-trivial automorphisms of $H$ explicitly, they are $$a \mapsto a^2,a \mapsto a^3,a \mapsto a^4$$
with orders $4,4,2$ respectively. There is thus a single valid homomorphism $\varphi$ determined by $$\varphi(x)(a)=a^4=a^{-1}.$$ There is therefore a unique non-abelian group of order 10, which admits the presentation
$$\langle a,x|a^5=x^2=1,xax^{-1}=a^{-1} \rangle$$
Since $D_{10}$ is a non-abelian group of order 10, it must be the unique non-abelian isomorphism type.
A: Here's another way of solving this problem, you could find useful:
$o(G)=$2·5
So using Sylow's Theorems we get $n_5$=1 and there are two options:


*

*$n_2$=1 $\implies$G$\cong$$\mathbb{Z}_{10}$

*$n_2$=5. In this case we have 5 elements of order 2 and just one of order 5 it's easy to check that $G\cong$$D_5$ 
