How to find the number of non-isomorphic groups of order 10? How to find the number of non-isomorphic groups of order 10?
Using Cauchy I can say that it has an element of order 2 and  an element of order 5,and so one group that I can manage is $\mathbb Z_{10}$
But how to make sure that if there are others also?Any help would be appreciated.
please dont give the answer only.Please provide the required details
In general if $G$ is a group having order $o(G)=p_1p_2p_3...p_n$
Is there a way to find the number of non-isomorphic groups of this order
 A: Up to isomorphism, there are only two groups of order $10$: the cyclic group $\mathbb{Z}/10\mathbb{Z}$, and the dihedral group $D_{10}$.
One way to see this is to first notice, as you did, that a group $G$ of order $10$ has an element of order $5$ and another of order $2$; the subgroups they generate, say $G_5$ and $G_2$, have order $5$ and $2$, respectively.
Now, since $G_5$ has index $2$, it is a normal subgroup of $G$.  Moreover, $G_5$ and $G_2$ have trivial intersection, and $G_5G_2=G$ (this can be seen by looking at the cardinality of $G_5G_2$).  Thus $G$ is the semidirect product of those two subgroups: $G=G_5\rtimes G_2$.
Now, semidirect products of the form $G_5\rtimes G_2$ are determined by an action of $G_2$ on $G_5$, or equivalently, by a homomorphism of groups $G_2\to Aut(G_5)$.  Since $G_5 \cong \mathbb{Z}/5\mathbb{Z}$, we have $Aut(G_5)\cong \mathbb{Z}/4\mathbb{Z}$, so there are exactly two group homomorphisms $G_2\to Aut(G_5)$ (since $G_2 \cong \mathbb{Z}/2\mathbb{Z}$).
Thus there are exactly two semidirect products $G_5\rtimes G_2$: the one corresponding to the trivial action of $G_2$ on $G_5$ gives $\mathbb{Z}/10\mathbb{Z}$, and the only non-trivial action of $G_2$ on $G_5$ gives $D_{10}$.
For your second, more general question, this post seems to provide the answer.
