Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\langle a\rangle \cap \langle b\rangle = \{e\}$?

share|improve this question

4 Answers 4

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.

share|improve this answer

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.

share|improve this answer
    
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. –  B. 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.

share|improve this answer

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.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.