Suppose that $G$ is a group of order $2020$. Then show $G$ has a unique subgroup of order $505$. By the Sylow Theorems, $G$ has a unique subgroup of order $101$, call it $P_{101}$. Then $G / P_{101}$ is a group of order $20$ and by the Sylow Theorems, $G / P_{101}$ has a normal subgroup of order $5$, call it $H_{5}$. Let $\pi : G \rightarrow G / P_{101}$ be the canonical map. The $\pi^{-1}(H_{5}) = \overline{H_{5}}P_{101}$, where $\overline{H_{5}}$ is the set of the $5$ distinct coset representatives. My questions are:
- Can I use the correspondence theorem to say that $\overline{H_{5}}P_{101}$ is a group?
- How can I show that $\overline{H_{5}}P_{101}$ has order $505$. Is it that $\left| \overline{H_{5}}P_{101} \right| = \frac{|\overline{H_{5}}|\cdot |P_{101}|}{|\overline{H_{5}} \cap P_{101}|} = 505$?
- How do I show that $\overline{H_{5}} P_{101}$ is a unique subgroup of order $505$?
- Can I claim that $\overline{H_{5}} P_{101}$ is normal and why?
1.) The preimage of a subgroup under a group homomorphism is a subgroup, yes.
2.) Yes, your calculation is correct. Essentially what you've said is: cosets are disjoint, each coset has $101$ elements, and $\bar H_5P_{101}$ is the union of $5$ cosets, so it has $505$ elements.
3.) Another subgroup of order $505$, let's call it $K$, would contain a subgroup of order $101$, by the Sylow theorems. Since $G$ has only one such subgroup, we conclude $P_{101} \le K$. What happens to $K$ under $\pi\colon G \to G/P_{101}$?
4.) Well, we know that $H_5$ is normal in $G/P_{101}$, so for any $g \in G$ and $h \in H_5$, we have $\pi(g)h\pi(g)^{-1} \in H_5$. Can you use this calculation somehow to show directly that $\bar H_5P_{101}$ is normal? Alternatively, what does part 3) imply about $g(\bar H_5P_{101})g^{-1}$?