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How can i show that if $o(G)=21$ and if $G$ has only one subgroup of order $3$ and only one subgroup of order $7$, then show that $G$ is cyclic.

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  • $\begingroup$ suppose an element of order 3 and an element of order 7 multiply then we get an element of order 21 ,so the group will be cyclic because those element commut , $\endgroup$
    – user112919
    Dec 2 '13 at 16:07
  • $\begingroup$ Why do the elements commute? Although this is true, this is basically the crux of the question. $\endgroup$
    – user61527
    Dec 2 '13 at 19:30
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Hint: Let $H$ and $K$ be the unique subgroups of order 3 and 7, respectively. Since they are the only subgroups of a given order they must be normal (why?). Conclude that $H, K$ are normal subgroups that intersect trivially, and so their internal direct product $HK \simeq H \times K$.

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  • $\begingroup$ Exactly. See my comment to Matt's answer. $\endgroup$ Jun 6 '12 at 15:33
  • $\begingroup$ For any $a\in G$, the set $a^{-1}Ha$ is also a subgroup of $G$($\because$ If $H$ is a subgroup of $G$, then $x^{-1}Hx$ is also a subgroup of $G$ for $x\in G$. Also if $H=\{e, h_{1},..., h_{k}\}$ then $a^{-1}Ha=\{e, a^{-1}h_{1}a,...,a^{-1}h_{k}a\}$ and these are all distinct elements of $a^{-1}Ha$, i.e. $oH)=o(a^{-1}Ha).$ But if $H$ is the only subgroup of a given order. Hence $a^{-1}Ha=H$. In other words $H$ is a normal subgroup of $G$. Am i right? $\endgroup$
    – Kns
    Jun 6 '12 at 15:40
  • $\begingroup$ @Kns: Yeah!$$$$ $\endgroup$ Jun 6 '12 at 15:43
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Use the Sylow Theorems. There exist groups of order 21 that are not cyclic: check the semi-direct products. Consider a group such that $G=\{x,y \mid x^7=y^3=1, yx=x^2y\}$ which is not cyclic but of order 21.

But, since there are only one Sylow-3 subgroup and Sylow-7 subgroup, they are both normal. You are right, these p-subgroups are cyclic, but since they are also normal, the group isomorphic to the direct product $\mathbb{Z}/7 \times \mathbb{Z}/3$ which is cyclic.

Alternative explanation without Sylow: Consider the number of elements. Using Lagrange's Theorem, the order of each element divides the order of the group. Consider the elements that are not in either subgroup, what are their possible orders?

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  • $\begingroup$ Thanks for answer! But in our abstract algebra course we can't study about Sylow theorems. $\endgroup$
    – Kns
    Jun 6 '12 at 15:21
  • $\begingroup$ I have added a second explanation without the use of Sylow, but using the fact that the order of each element in $G$ divides the order of $G$ $\endgroup$ Jun 6 '12 at 15:27
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    $\begingroup$ One does not need the Sylow theorems to prove that if there is one one subgroup of a given order, it is normal. Consider: for any subgroup $H$ of a given order and $g \in G$, $gHg^{-1}$ is also a subgroup of the same order. Since there is only ONE subgroup of that order, these must coincide for any $g$, hence $H$ is normal. $\endgroup$ Jun 6 '12 at 15:30
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What are the possible orders of elements for this group? One of order 1, two of order 3 (can't have any more as that would mean another subgroup of order 3), six of order 7 (and likewise no more). What of the rest, what must they be?

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