Equivalence of definitions of Sylow subgroup of normal subgroup

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.

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) uses::Sylow subgroups exist
 * 2) uses::Sylow implies order-dominating
 * 3) uses::Product formula
 * 4) uses::Sylow satisfies intermediate subgroup condition

(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 $$P \le gTg^{-1}$$. Let $$S = gTg^{-1}$$. Then, $$P \le S$$. Also, $$P \le N$$, so $$P \le N \cap S$$.

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.

Textbook references

 * , Page 101, Exercise 9, Section 3.3, and Page 147, Exercise 34, Section 4.5 (Sylow's theorem)