I wanted to ask if I had done this problem correctly.
Let $G$ be a group of order $pqr$ (for $p > q > r$ primes).
(i) If $G$ fails to have a normal subgroup of order $p$, determine the number of elements with order $p$.
(ii) If $G$ fails to have a normal subgroup of order $q$, show that there are at least $q^2$ elements of order $q$.
(iii) Show that $G$ has a nontrivial normal subgroup.
Let $n_p$ be the number of Sylow $p$-groups. For (i), by hypothesis $n_p > 1$, so $n_p$ must be one of $r, q$ or $qr$. Since $p > r > q$ and $n_p \equiv 1 \pmod p$, we must must have $n_p = qr$. Every element besides the identity of a cyclic group of prime order is a generator, so these $qr$ groups have trivial intersection, meaning there are $(p-1)qr = pqr – qr$ elements of order $p$.
(ii) By hypothesis $n_q > 1$, so $n_q$ can either be $p, r,$ or $pr$. Clearly $n_q$ can't be $r$, so at the very least there are $p$ Sylow $q$-groups, and therefore at least $(q-1)p$ elements of order $q$. Since $p > q > r$, $p$ and $q$ must be odd primes, and therefore $p = q + k$ with $k \geq 2$. Thus $$(q-1)p = (q-1)(q+k) = q^2 + kq – q + k \geq q^2 + 2q – q + 2 > q^2$$ (iii) If $n_r, n_p,$ or $n_q = 1$ we are done. So assume all of these numbers are $> 1$. By parts (i) and (ii) we can find at least $pqr – qr$ elements of order $p$, and $q^2$ elements of order $q$.
The elements which do not have order $p$ or $q$ must be numbered $|G| – (pqr – qr) – q^2 = qr – q^2 = q(r-q)$ or less. But this number is negative, contradiction.
Best Answer
For the third part, we have: Let $p$ be the smallest prime dividing the order of $G$. If $G$ has a subgroup $H$ of with $[G:H]=p$, then $H$ is normal. Of course, firstly, you should verify this statement.(Normal subgroup of prime index)
In this question, $\vert$$G$$\vert=pqr$ and $r$ is the smallest prime dividing $\vert$$G$$\vert$. For the subgroup of order $pq$ ; that is , of index $r$, you may consider the Sylow-$p$-subgroups, Sylow-$q$-subgroups and their products.