Only two groups of order $10$: $C_{10}$ and $D_{10}$

Question

Show that up to isomorphism there exist only two groups of order 10: $C_{10}$ and $D_{10}$. I need some help on this question. I only know the basic definitions of isomorphism. Any hints?

Answer 1

A group $G$ of order $10$ has a subgroup $H$ of order $5$ (which must be cyclic). As it is of index $2$, $H$ is normal in $G$. If $g\in G$ is of order $2$, then the conjugation $x\mapsto gxg^{-1}$ is an automorphims of $H$. As $C_5$ has only two automorphisms and the rest of the group structure is determined by these observations, there are no other groups of order $10$.

Answer 2

Hint:

$G$ has an element $y$ of order $5$ and an element $x$ of order $2$.

The sets $e, y, y^2, y^3, y^4$ and $ x, xy, xy^2, xy^3, xy^4$ are disjoint. Therefore these are all 10 elements in the group.

Now $yx$ must be one of these elements.

Show that $yx$ cannot be one of $e, y, y^2, y^3, y^4, x$.v So it can be $xy, xy^2, xy^3, xy^4$.

If $yx=xy$ then show that you have $C_{10}$.

If $yx=xy^{4}$ then you get $D_{10}$.

If $yx=xy^2$ then you get $y=xy^2x$ and $y^2=xyx$. Show that those contradict eachother.

If $yx=xy^3$ get a contradiction.

Answer 3

More generally, take $p,q$ primes. I claim there exist at most two groups of order $pq$ for any choice of $(p,q)$. Indeed, suppose without loss of generality that $p<q$, and let $G$ be a group of order $p,q$. By Cauchy there exist in $G$ an element $a$ of order $p$ and an element $b$ of order $q$. Since $p$ is the least prime dividing the order of $G$, the cyclic group generated by $b$ (which has order $q$) has index $p$ in $G$, so $\langle b\rangle \lhd G$. Since $p,q$ are prime, $\langle a\rangle \cap \langle b\rangle $ is trivial. It follows that $G$ is generated by $a,b$, that is $G=\langle a,b\rangle$.

To understand $G$ it then suffices to understand how $a$ acts on $b$, and since $b$ is normal we know that $aba^{-1}$ is a power of $b$, say $aba^{-1}=b^j$. Since $a^p=1$, we get $b=a^pba^{-p}=b^{j^p}$ or what is the same, that $j^p=1\mod q$. This means there is an element of order dividing $p$ in the group of units modulo $q$, so that either $j=1$ and $a$ and $b$ commute so $G$ is the group $C_p\times C_q=C_{pq}$, or $j$ has order exactly $p$ and $p\mid q-1$.

It suffices you check that if $j$ has order exactly $p$ then the group with presentation $$\langle a,b\mid a^p,b^q,aba^{-1}=b^j\rangle$$ doesn't depend on the choice of $j$. I leave that to you. For example, when you take $p=2, q$ any prime you either get $C_{2q}$ or $D_{2q}$ which is what you want. Indeed, the equation $j^2=1\mod q$ has solutions $j=1,-1$ and this gives either $$\langle a,b\mid a^2,b^q,ab=ba\rangle \simeq C_2\times C_q$$ or $$\langle a,b\mid a^2,b^q,aba^{-1}=b^{-1}\rangle \simeq D_{2q}=C_2\rtimes C_q$$