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This the proof we were given as to why a group of order 143 is cyclic.

Let $G$ be a group of order 143. By the Sylow Third Thm., the # of Sylow 11-subgroups of $G$ is $1+11k$ and is a factor of 13, so $k=0$ and there is an element $a\in G$ such that $\mathrm{Ord}(a) = 11$ and $\langle a\rangle$ is normal to $G$. Similarly, the number of Sylow 13-subgroups of $G$ is $1+13k$ and is a factor of 11, so $k=0$ and there is an element $b\in G$ such that $\mathrm{Ord}(b)=13$ and $\langle b\rangle$ is normal to $G$. Since $\langle a\rangle \cap\langle b\rangle = \{e\}$ we see that $ab=ba$, it follows that $\mathrm{Ord}(ab) =143$.

My question is how do we know that $\langle a\rangle $ and $\langle b\rangle$ are normal to G? What allows us to say that $\langle a\rangle \cap \langle b\rangle = \{e\}$?

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up vote 4 down vote accepted

Sylow's second theorem states that every Sylow $p$-subgroup of a group $G$ is conjugate to the others; that is, for any Sylow $p$-subgroups $A,B\subset G$, there is some $g\in G$ such that $B=gAg^{-1}$.

It is also easy to see that any subgroup of $G$ that is conjugate to a Sylow $p$-subgroup must also be a Sylow $p$-subgroup; this is because conjugation is a bijection, and the Sylow $p$-subgroups are defined by the property of having order $p^n$, where $p^n$ is the largest power of $p$ dividing $|G|$.

Thus, if you know that there is a unique Sylow $p$-subgroup $H\subset G$ for a given $p$, then for any $g\in G$, the conjugate subgroup $gHg^{-1}$ is also a Sylow $p$-subgroup of $G$, and therefore must be equal to $H$; thus $gHg^{-1}=H$ for all $g\in G$, i.e. $H$ is normal.

This is mentioned as one of the important consequences of the Sylow theorems in the Wikipedia article.

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Every Sylow $p$-subgroup of a finite group $G$ is conjugate to each other. Thus, if there is a unique Sylow $p$-subgroup, then it must equal all of its conjugates (since all of its conjugates have the same order and hence are also Sylow $p$-subgroups), and hence it is a normal subgroup.

If $H,K$ are subgroups of a finite group $G$ with coprime orders, then suppose there is a nontrivial element $x\in H\cap K\setminus\{1_G\}$. Since $x\in H$, its order $|x|$ (the size of the cyclic subgroup it generates) divides $|H|$, by Lagrange's theorem, and similarly $x\in K$ implies $|x|$ divides $|K|$. Since $\gcd(|H|,|K|)$ is unity, $|x|=1$ and hence $x=1_G$, a contradiction. Hence $H\cap K$ is trivial.

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What do you mean by "unity"? – Low Scores Dec 21 '12 at 5:59
@LowScores Sorry, it's a word for "one." I don't like starting new lines with a numeral, one of my many OCDish things... – anon Dec 21 '12 at 6:03
+1 for nice illustration. – Babak S. Dec 21 '12 at 8:34

By the third Sylow theorem, we know that the number of Sylow 11-groups is 1 mod 11. We also know that they are all conjugate, so the orbit-stabilizer theorem tells us that the number of Sylow 11-groups divides the order of the group. Because $143=11\times 13$, there can only be one. No other factors are $1$ mod $11$.

A similar argument works for the $13$-group.

Now let the whole group act on the $11$-group by conjugation. Note that being normal is equivalent to having exactly $1$ conjugate under this action. This is what we showed above.

The final part is just the direct product theorem for groups.

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Also you can use this fact that, every cyclic group of order $p$ is isomorphic to $\mathbb Z_p$. So, since $11|143,\;\; 13|143$ and what @anon noted completely in the second paragraph, we have a group of order $11$ and $13$ such that $\{1\}$ is their intersection. and $G\cong\mathbb Z_{11}\times\mathbb Z_{13}\cong\mathbb Z_{11\times 13}$. Note that since $G$ is the direct product of its $p-$ sylow subgroups, it is nilpotent also. However; this additional fact can be arisen for being cyclic as well.

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Nice observations, and to the point, my dear friend! +1 – amWhy Apr 14 '13 at 1:29

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