2
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ Do you assume that $p_1, \ldots, p_n$ are pairwise distinct primes? $\endgroup$ Commented Feb 17, 2015 at 15:35

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ "this post" provides an answer for $o(G)=30$ I was asking in general $\endgroup$
    – Learnmore
    Commented Feb 18, 2015 at 2:40
  • 1
    $\begingroup$ @learnmore Even though the question in the link was about groups of order $30$, the answer I linked to provides a reference for all groups with cyclic Sylow subgroups - in particular, if the order of $G$ is $p_1\ldots p_n$ with the $p_i$ pairwise distinct, then the post does provide an answer to your question. $\endgroup$ Commented Feb 19, 2015 at 11:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .