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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?

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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 $k$.
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  • $\begingroup$ I have got the first two bulletins ;I can't get the 3rd one please help sir $\endgroup$ – Learnmore Mar 2 '15 at 11:22
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    $\begingroup$ Assume that $x^{-1}ax = a^2$. By squaring you get $x^{-1}a^2x = a^4$. Conjugate both sides by $x$, and you'll get $a^2 = x^{-1}a^4x$. Thus $x^{-1}ax = x^{-1}a^4x$, hence $a = a^4$, a contradiction. The other cases are similar. $\endgroup$ – Leppala Mar 2 '15 at 11:53
  • $\begingroup$ @learnmore: Leppälä's way or combine $$a=1a1^{-1}=b^2ab^{-2}=a^{k^2}$$ with $a^5=1$. $\endgroup$ – Jyrki Lahtonen Mar 2 '15 at 11:58
  • $\begingroup$ sir i have got your point $b$ has order two so second point gives $a=a^{k^2}$ i.e $k^2-1=0 $ so $k^2=1$ hence only possibility is $b^2a=ab^2$ is this correct? $\endgroup$ – Learnmore Mar 2 '15 at 12:14
  • $\begingroup$ Not quite there yet, @learnmore. Remember that A) you ruled out $k=1$, because that leads to an abelian group, B) because $a$ is of order five, we have that $$a^i=a^j\Leftrightarrow i\equiv j\pmod 5.$$ So you only get $k^2-1\equiv0\pmod5$. $\endgroup$ – Jyrki Lahtonen Mar 2 '15 at 12:33
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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.

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  • $\begingroup$ what is semi direct product ? $\endgroup$ – Learnmore Mar 2 '15 at 11:08
  • $\begingroup$ I have learnt direct product $\endgroup$ – Learnmore Mar 2 '15 at 11:08
  • $\begingroup$ Semi-direct products are a construction similar but more general than the direct products. Some groups can be represented as a semi-direct product of two smaller groups, which inductively helps us classifying groups of a given order. The definition can be found here $\endgroup$ – user1337 Mar 2 '15 at 11:12
  • $\begingroup$ which book should I consult it for sir?@ user 1337 $\endgroup$ – Learnmore Mar 2 '15 at 11:21
  • $\begingroup$ Personally I used Dummit & Foote's Abstract Algebra 3rd edition. It's on chapter 5. $\endgroup$ – user1337 Mar 2 '15 at 11:25
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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:

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

  2. $n_2$=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$

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