Equivalence of definitions of Sylow subgroup of normal subgroup

This article gives a proof/explanation of the equivalence of multiple definitions for the term Sylow subgroup of normal subgroup
View a complete list of pages giving proofs of equivalence of definitions

The definitions that we have to prove as equivalent

The following are equivalent for a subgroup of a finite group:

1. It is a Sylow subgroup of a normal subgroup of the whole group.
2. It is the intersection of a normal subgroup of the whole group with a Sylow subgroup of the whole group.
3. It is a Sylow subgroup inside its normal closure.

Related facts

Similar facts

• Intersection of Hall subgroup and permuting subgroup is Hall in that subgroup
• Intersection of Hall subgroup and normal subgroup implies Hall subgroup of normal subgroup
• Equivalence of definitions of Sylow subgroup of permutable subgroup

Facts involving normal Sylow subgroups instead

• Normal Sylow satisfies transfer condition: This states that the intersection of a normal Sylow subgroup with any subgroup is normal Sylow in the other subgroup.
• Normal Hall satisfies transfer condition

Facts used

1. Sylow subgroups exist
2. Sylow implies order-dominating
3. Product formula
4. Sylow satisfies intermediate subgroup condition

Proof

(1) implies (2)

Given: A group $G$, a normal subgroup $N$ of $G$, a $p$-Sylow subgroup $P$ of $N$.

To prove: There exists a $p$-Sylow subgroup $S$ of $G$, such that $S\cap N=P$.

Proof: By fact (1), there exists a $p$-Sylow subgroup $T$ of $G$, and by fact (2), there exists $g\in G$ such that $P\leq gTg^{-1}$. Let $S=gTg^{-1}$. Then, $P\leq S$. Also, $P\leq N$, so $P\leq N\cap S$.

On the other hand, since $S$ is a $p$-group, so is $N\cap S$. But $P$ is a $p$-subgroup of the largest possible order in $N$, forcing $P=N\cap S$.

(2) implies (1)

Given: A group $G$, a normal subgroup $N$ of $G$, a Sylow subgroup $S$ of $G$.

To prove: $S\cap N$ is a Sylow subgroup of $N$.

Proof: By the product formula (fact (3)), we have:

${\frac {|NS|}{|S|}}={\frac {|N|}{|N\cap S|}}$

Since $N$ is a normal subgroup, $N$ and $S$ are permuting subgroups, so $NS$ is a subgroup. In particular, $|NS|/|S|$ is a divisor of $|G|/|S|$, and is hence relatively prime to $p$. Thus, the right side is also relatively prime to $p$. So, $N\cap S$ is a $p$-subgroup of $N$ whose index is relatively prime to $p$, and hence must be a $p$-Sylow subgroup.

Equivalence of (1) and (3)

Clearly (3) implies (1). For the converse, use fact (4) (a Sylow subgroup in the whole group is also a Sylow subgroup in any intermediate subgroup). For a general statement to this effect, refer: Composition of subgroup property satisfying intermediate subgroup condition with normality equals property in normal closure.

References

Textbook references

• Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, ^{More info}, Page 101, Exercise 9, Section 3.3, and Page 147, Exercise 34, Section 4.5 (Sylow's theorem)