Cyclicity from Sylow Subgroups

After classifying some small groups, we now focus on a recurring method: prove that enough Sylow subgroups are unique, then assemble the whole group from them. In this lesson, we focus in the cyclicity from Sylow subgroups. The key point is that normal subgroups of relatively prime orders commute elementwise when their intersection is trivial. This turns Sylow information into a direct product decomposition and often proves that the group is cyclic.

Let

G

be a finite group. We say that

G

has \textbf{cyclicity from Sylow subgroups} when its Sylow subgroups are normal, cyclic, have pairwise relatively prime orders, and multiply to the whole group.

Let

G

be a finite group. We say that

G

has \textbf{cyclicity from Sylow subgroups} when its Sylow subgroups are normal, cyclic, have pairwise relatively prime orders, and multiply to the whole group.

Let

H

and

K

be normal subgroups of a group

G

. If

H\cap K=\{e\}

, then every element of

H

commutes with every element of

K

Given that

H\trianglelefteq G

K\trianglelefteq G

, and

H\cap K=\{e\}

. To prove that every element of

H

commutes with every element of

K

. Let

h\in H

and

k\in K

. Consider the commutator

\begin{equation*} hkh^{-1}k^{-1}. \end{equation*}

Since

K

is normal,

hkh^{-1}\in K

, so

hkh^{-1}k^{-1}\in K

. Since

H

is normal,

kh^{-1}k^{-1}\in H

, and therefore

hkh^{-1}k^{-1}\in H

. Thus

\begin{equation*} hkh^{-1}k^{-1}\in H\cap K=\{e\}. \end{equation*}

Therefore

\begin{equation*} hkh^{-1}k^{-1}=e. \end{equation*}

This gives

\begin{equation*} hk=kh. \end{equation*}

Hence every element of

H

commutes with every element of

K

\square

Let

G

be a finite group whose Sylow subgroups are all normal and cyclic, with pairwise relatively prime orders. If their product has order

|G|

, then

G

is cyclic.

Given that the Sylow subgroups of

G

are normal and cyclic, have pairwise relatively prime orders, and their product has order

|G|

. To prove that

G

is cyclic. Let the Sylow subgroups be

P_1,P_2,\ldots,P_k

. Since their orders are pairwise relatively prime, we have

\begin{equation*} P_i\cap P_j=\{e\} \end{equation*}

whenever

i\neq j

. By the preceding lemma, elements from distinct

P_i

commute elementwise. Therefore the product is an internal direct product:

\begin{equation*} G=P_1P_2\cdots P_k\cong P_1\times P_2\times\cdots\times P_k. \end{equation*}

Each

P_i

is cyclic, and their orders are pairwise relatively prime. A direct product of cyclic groups of pairwise relatively prime orders is cyclic. Hence

G

is cyclic.

\square

This preview shows how normal Sylow subgroups can assemble into a cyclic direct product. Enter a group order or choose a preset from the solved problems. The display lists the Sylow factor orders and checks that they are pairwise coprime. When the corresponding Sylow subgroups are normal and cyclic, these factors commute and their product is cyclic.

The preview identifies the Sylow factor orders but does not prove normality by itself. The written examples use Sylow counting first, then the direct product argument turns the normal cyclic factors into a cyclic subgroup or cyclic group.

This calculator checks the direct-product part of the argument. Enter factor orders such as `11,7`, `5,7,19`, or `3,41`. The calculator tests whether the orders are pairwise coprime and whether their product has the expected order. When the factors are cyclic and normal in the surrounding group, the direct product is cyclic.

The calculator isolates the final algebraic step. Sylow theory supplies normality and subgroup existence; coprime cyclic direct products supply cyclicity.

Let

G

be a group of order

231=3\cdot 7\cdot 11

. Show that

G

has a cyclic subgroup of order

77

Let

H

be a Sylow

11

-subgroup of

G

. Then

\begin{align*} n_{11} &\equiv 1 \pmod {11}, \\ n_{11} &\mid 21. \end{align*}

The divisors of

21

are

1,3,7,21

, and only

1

is congruent to

1

modulo

11

. Hence

n_{11}=1

, so

H\trianglelefteq G

. Let

K

be a Sylow

7

-subgroup. Then

\begin{align*} n_7 &\equiv 1 \pmod 7, \\ n_7 &\mid 33. \end{align*}

The divisors of

33

are

1,3,11,33

, and only

1

is congruent to

1

modulo

7

. Hence

n_7=1

, so

K\trianglelefteq G

. Since

|H|=11

and

|K|=7

, we have

H\cap K=\{e\}

. Normality gives elementwise commutativity. Thus

\begin{equation*} HK\cong H\times K\cong C_{11}\times C_7\cong C_{77}. \end{equation*}

Hence

G

has a cyclic subgroup of order

77

\square

Let

G

be a group of order

5\cdot 7\cdot 19

. Prove that

G

is cyclic.

Let

|G|=5\cdot 7\cdot 19

. Let

n_{19}

be the number of Sylow

19

-subgroups. Then

\begin{align*} n_{19} &\equiv 1 \pmod {19}, \\ n_{19} &\mid 35. \end{align*}

Thus

n_{19}=1

. Let

n_7

be the number of Sylow

7

-subgroups. Then

\begin{align*} n_7 &\equiv 1 \pmod 7, \\ n_7 &\mid 95. \end{align*}

The divisors of

95

are

1,5,19,95

, and only

1

is congruent to

1

modulo

7

. Hence

n_7=1

. Let

n_5

be the number of Sylow

5

-subgroups. Then

\begin{align*} n_5 &\equiv 1 \pmod 5, \\ n_5 &\mid 133. \end{align*}

The divisors of

133

are

1,7,19,133

, and only

1

is congruent to

1

modulo

5

. Hence

n_5=1

. Thus all Sylow subgroups are normal. Their orders are pairwise relatively prime and each is cyclic because it has prime order. Therefore

\begin{equation*} G\cong C_5\times C_7\times C_{19}\cong C_{5\cdot 7\cdot 19}. \end{equation*}

Hence

G

is cyclic.

\square

Let

G

be a group of order

123

. Show that for every positive divisor

d

123

, there is a unique subgroup of

G

of order

d

Since

\begin{equation*} 123=3\cdot 41 \end{equation*}

and

\begin{equation*} 3\nmid 40, \end{equation*}

every group of order

123

is cyclic by the cyclic criterion for groups of order

pq

. Hence

\begin{equation*} G\cong C_{123}. \end{equation*}

A cyclic group has a unique subgroup of order

d

for every positive divisor

d

of its order. Therefore

G

has a unique subgroup of order

d

for every positive divisor

d

123

\square

1. Show that every group of order

133

is cyclic. 2. Let

G

be a group of order

70

. Prove that

G

has a cyclic subgroup of order

35

. 3. Explain why normal Sylow subgroups of relatively prime orders commute elementwise.

1. Since

133=7\cdot 19

and

7\nmid 18

, every group of order

133

is cyclic. 2. The Sylow

7

- and Sylow

5

-subgroups are unique and normal. Their product is isomorphic to

C_7\times C_5\cong C_{35}

. 3. If

H

and

K

are normal with

H\cap K=\{e\}

, then each commutator

hkh^{-1}k^{-1}

lies in both

H

and

K

, so it equals

e

Normal Sylow subgroups of relatively prime orders commute and form a direct product; if each factor is cyclic and the orders are coprime, the product is cyclic.

No. The orders must be pairwise relatively prime for the direct product to be cyclic.

Every group of prime order is cyclic.

BMLabs Mathematics

BMLabs Mathematics

Cyclicity from Sylow Subgroups

Cyclicity from Sylow Subgroups

CYCLICITY FROM SYLOW SUBGROUPS

EXPLORE NORMAL SYLOW PRODUCTS

NORMAL SYLOW PRODUCT BUILDER

INTERPRET THE PRODUCT BUILDER

CALCULATE CYCLIC PRODUCTS FROM SYLOW FACTORS

CYCLIC PRODUCT FROM SYLOW FACTORS CALCULATOR

CONNECT THE CALCULATION

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