TITLE: Group of order $40$
QUESTION [5 upvotes]: I need to show that any group $G$ such that $|G|=40$ has 3 subgroups $H_1,H_2,H_3$ such that $H_1<H_2<H_3$ and their orders are $5,10,20$ accordingly.
Thanks a million!

REPLY [2 votes]: You know that the group has a unique (and thus normal) $5$-Sylow subgroup, so call this $H_1$.
You also know that the group has at least one $2$-Sylow subgroup, which will have order $8$. Now, this also means that the group will have subgroups of order $2$ and $4$ (see for example A group of order $8$ has a subgroup of order $4$).
Let $A$ be a subgroup of order $2$ and $B$ be a subgroup of order $4$ with $A\leq B$. Now since $H_1$ is normal, both $H_2 = H_1A$ and $H_3 = H_1B$ are subgroups, and it is easy to check that they satisfy the requirements.