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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?

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  • $\begingroup$ 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
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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.

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