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I am trying to classify all groups of order 10 upto isomorphism. Suppose $G$=10. Using Sylow's theorem, if $n_p$ is the number of the p-Sylow groups (p prime), then $n_2|5$, so it is 1 or 5. Similarly,$n_5$ is 1 or 2. But $n_5\equiv 1 \bmod $, so $n_5$ can only be 1.

Case I: $n_5$ is 1. Then given $d\in \mathbb{Z}$, the group has at most 1 subgroup of order $d$, so $G$ is cyclic. So, $G\cong \mathbb{Z_{10}}$ .

Case II: This is where I am stuck. We have 5 subgroups of order 2 and one subgroup of order 5 (call it H). Since $|H|$ is prime, I infer $H$ is a cyclic group of the form $\{1,r,\dots,r^4\}$. Also, $H$ is a unique subgroup of prime order in G, so $H\unlhd G$. But I don't have an idea how to proceed after this. Thanks.

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  • $\begingroup$ Count elements - you have the identity, five of order two and four of order five. There are no more to find. Now take an element $a$ of order $2$ and examine the coset $aH=Ha$ $\endgroup$ – Mark Bennet Oct 24 '15 at 13:21
  • $\begingroup$ Case I does not mean that $G$ is cyclic. You need that true for all $d\in\mathbb Z$ to conclude, and you only have it for $d=5$. Perhaps you meant $n_2=1$? $\endgroup$ – Thomas Andrews Oct 24 '15 at 13:24
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Consider it this way. Let $a$ be an element of order $2$ and $b$ an element of order $5$. Show that $a^ib^j$ is all the elements of the group, with $i=0,1$ and $j=0,1,2,3,4$. Then $aba^{-1}$ is of order $5$, so it is equal to $b^j$ for some $j$. But then $b=ab^{j}a^{-1}=(aba^{-1})^j=b^{j^2}$. So $j^2\equiv 1\pmod 5$. What are possible values for $j$?

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  • $\begingroup$ Thank you very much. So, using your hint we can show that $j$ is either 4 or 1. If $j$ is 1, then we have $ab=ba$. In class, I have proved that if $a$ and $b$ commute with $\gcd (o(a),o(b))=1$, then $ab$ has order 10 which proves that $G$ is cyclic. If $j$ is 4, $aba^{-1}=b^4$ i.e. $aba=b^4a^2=b^4=b^{-1}$ which means $G$ is $D_{10}$. $\endgroup$ – user96343 Oct 25 '15 at 17:36
  • $\begingroup$ I have just one question. What was your motivation for looking at $aba^{-1}$? $\endgroup$ – user96343 Oct 25 '15 at 17:39
  • $\begingroup$ Yes, to your first comment, that is correct. On the second: In general, expression of the form $aba^{-1}$ are called "conjugates" of $b$, and conjugates of $b$ tend to "behave like $b$." In particular, the map: $g\to aga^{-1}$ is an isomorphism of the group $G$ with itself. Also, the subgroup generated by $b$ must be normal, so this map actually sends that subgroup to itself. Basically, it is often the case when studying groups that we look at conjugates. $\endgroup$ – Thomas Andrews Oct 25 '15 at 17:55
  • $\begingroup$ Another motivation is that you want $ba=a^ib^j$ for $i=0,1$ $j=0,1,2,3,4$. But we can quickly show that $i\neq 0$, because if $ba=b^{j}$ then $a=b^{j-1}$. So $ba=ab^j$ for some $j$, which gets us $aba=b^j$. $\endgroup$ – Thomas Andrews Oct 25 '15 at 18:02
  • $\begingroup$ okay, got it. thanks. $\endgroup$ – user96343 Oct 25 '15 at 18:17

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