From artin's 2nd edition of Algebra on the sylow theorem section.

Problem: Classify groups of order (a) 33

Answer: Let $G$ be the group. $n(G)=33$

Factors of $33$ are $1,3,11,33$

If for any any element $g, \sigma(g) = 1, 3, 11$ or 33 g is cyclic

Consider $g$ of order 11.

By the sylow theorem the number of subgroups of order $11 = n_{11} = 1 + 11k$

and divides 3 $ \rightarrow k = 0 \rightarrow n_{11} = 1 \rightarrow K$ is the only subgroup of order 11 ==>

K is normal ==> G/K is a group of order 3 ==> G/K = is also cyclic ==> h3 = gi.

If i > 0 then ord(h) = 3*11 ==> G is cyclic. If i = 0 the h3 = 1, and H = is s subgroup of order 3.

By Sylow theorem, the number of subgroups of order 3 = n_3 = 1 + 3k and divides 11

$\rightarrow k = 0 \rightarrow n_{3} = 1 \rightarrow$

H is the only subgroup of order 3

This implies that H is normal.

$H$ and $K$ are of orders 3 and 11 which are coprime to each other with common factor 1.

$hgh^{-1}g^{-1} = (hgh-1)g-1 = h(gh-1g-1)$ belong to both $H$ and $K$ hence has to be $1$.

This implies $hg = gh$ and now since $g$ and $h$ commute and relatively prime

$\sigma(h) \sigma(g) = \sigma(hg) = 3 \times 11 =33$

Or $hg$ generates G

Or $G$ is cyclic.

Is this correct?

  • $\begingroup$ what is your question? $\endgroup$ – Kushal Bhuyan Dec 6 '15 at 16:32
  • $\begingroup$ $|G|=3\times 11$ so it is abelian and cyclic. Has two normal subgroup of order $11$ and $3$ respectively. $\endgroup$ – Kushal Bhuyan Dec 6 '15 at 16:33

This can be proved without an appeal to the Sylow theorems.

By Cauchy's theorem, $G$ has an element of order $11$, let $Q$ be the subgroup generated by this element. Hence $|Q|=11$. Observe that $Q$ has to be normal. For if $Q^*$ is a conjugate of $Q$ and $Q \neq Q^*$, then $Q \cap Q^*=1$, since $11$ is prime. But then $|QQ^*|=\frac{|Q||Q^*|}{|Q \cap Q^*|}=11^2 \gt |G|$, which is absurd.

Since $Q$ is abelian, we have $Q \subseteq C_G(Q) \subseteq N_G(Q)=G$. But $|G|=3 \cdot 11$, so $|G:C_G(Q)|$ equals $1$ or $3$. In the latter case, we have $Q=C_G(Q)$, and we apply the $N/C$ theorem: $G/Q$ embeds homomorphically in $Aut(Q) \cong C_{10}$. However, this obstructs $3$ not dividing $10$.

We conclude that $G=C_G(Q)$, which is equivalent to $Q \subseteq Z(G)$, the center of $G$.
Finally, again by Cauchy's theorem we can find a subgroup $P$ of order $3$. But then $|PQ|=\frac{|P||Q|}{|P \cap Q|}=33$, so $G=PQ$, and since $Q$ is central, $P$ is certainly normal in $G$.We now have $G=PQ$, $P$ and $Q$ are both normal, and $P \cap Q=1$. It follows that $G \cong P \times Q$ and we are done.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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