# Normal Sylow satisfies transfer condition

## Statement

### Statement with symbols

Suppose $H$ is a normal Sylow subgroup of a finite group $G$ and $K \le G$ is any subgroup. Then $H \cap K$ is a normal Sylow subgroup of $K$.

## Related facts

### Intersecting a normal subgroup with a Sylow subgroup

• Equivalence of definitions of Sylow subgroup of normal subgroup: This states that a subgroup that is the intersection of a Sylow subgroup and a normal subgroup is a Sylow subgroup of the normal subgroup. Conversely, a subgroup expressible as a Sylow subgroup of a normal subgroup is expressible as an intersection of the normal subgroup with a Sylow subgroup of the whole group.

## Proof

Given: A finite group $G$, a normal Sylow subgroup $H$, and a subgroup $K \le G$.

To prove: $H \cap K$ is a normal Sylow subgroup of $K$.

Proof: By the second isomorphism theorem, $H \cap K$ is normal in $K$, and we have:

$K/(H \cap K) \cong HK/H$.

Here, $HK$ is a subgroup of $G$. Since $G/H$ has order relatively prime to $p$, so does $HK/H$ (by fact (2)). Thus, $K/(H \cap K)$ has order relatively prime to $p$, so $[K:H \cap K]$ is relatively prime to $p$. Thus, $H \cap K$ is a normal $p$-subgroup of $K$ with index relatively prime to $p$, and is thus a normal $p$-Sylow subgroup of $K$.

## 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 147, Exercise 32, Section 4.5 (Sylow's theorem)