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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?

Olexandr Konovalov
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Lin
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  • Can you show that such a group has a subgroup of order $5$ and that it is normal? – Travis Willse Feb 03 '16 at 18:11
  • Let A and B be two normal subgroups. I proved previously if A intersect B is the identity element and G=AB then G is isomorphic to AxB.Is this of any help? – Lin Feb 03 '16 at 18:20
  • Problem 55 can help lead you in the direction of the proof of a more general result http://abstract.ups.edu/aata/exercises-isomorph.html – eepperly16 Feb 03 '16 at 18:25
  • Watch this https://math.stackexchange.com/questions/291349/looking-for-a-simple-proof-that-groups-of-order-2p-are-up-to-isomorphism-mat – x100c Dec 29 '20 at 17:08

3 Answers3

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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.

N. S.
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  • This is a seriously underrated answer! Such insight would have been quite useful when I was studying algebra. – The Chaz 2.0 Feb 04 '16 at 22:37
  • Why must $G$ have an element of order $5$ and an element of order $2$? Lagrange surely doesn't guarantee this. – Irregular User Feb 07 '16 at 22:54
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    @IrregularUser https://en.wikipedia.org/wiki/Cauchy's_theorem_%28group_theory%29 – N. S. Feb 07 '16 at 23:28
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    @IrregularUser In this case one can prove the existence of those elements at an elementary level, no Cauchy Theorem is needed. By Lagrange every element has order 1,2,5 or 10, and there is an element of order 1. If there is an element of order 10 we are done. Otehrwise every element has order 2 or 5.... The elements of order 5 can be grouped in groups of $4$ (as together with $e$ they generated a 5-group). Thus it is impossible for the remaining $9$ elements to all have order $5$. So either there is an element of order 5 and one element of order 2, or all elements have order 2... – N. S. Feb 07 '16 at 23:32
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    ... The last is impossible: in that case the group would be Abelian (standard exercise in any introductory book) and it is not too hard to see that it is impossible for an Abelian group to have 10 elements, and all elements to have order 1 or 2... – N. S. Feb 07 '16 at 23:34
  • @N.S. Why can't an abelian group have an order of 10? – 4M4D3U5 M0Z4RT Feb 05 '20 at 10:48
  • @4M4D3U5M0Z4RT $C_{10}$ is cyclic, thus Abelian. – N. S. Feb 05 '20 at 14:19
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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 $10$.

Hagen von Eitzen
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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 $$\langle a,b\mid a^2,b^q,aba^{-1}=b^{-1}\rangle \simeq D_{2q}=C_2\rtimes C_q$$

Pedro
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