Conditions for cyclic quotient group

Let $G$ be an arbitrary finite group and $H$ a normal subgroup.

What are some good conditions on $H$ that make the quotient $G/H$ cyclic?

I want to avoid any further restriction on $G$.

Solution 1:

This is largely the wrong question. Let $H,K$ be groups and let $G = H \times K$ so that $H \trianglelefteq G$. (This is not the only way to get to the upcoming punchline.)

What sort of conditions can you apply to $H$ to force $K \cong G/H$ cyclic? (Notice that $K$ was chosen entirely independently of $H$.)

There are no such conditions generally. You say you don't want to constrain $G$, but $G$ is the only thing that controls the cosets of $H$. $H$ is stuck in one of its cosets and has no control over the rest of $G$.

Notice that no one is giving you constraints on $H$. They constrain the number of cosets of $H$ (via $|G|/|H|$) or they require $G$ special (see especially the finite version of special). This is because constraining a tiny chunk of $G$ is too weak to control the cosets of that chunk.

Solution 2:

Another sufficient condition that captures some of those already mentioned is that $\gcd([G:H], \phi([G:H]))= 1$, where $\phi$ is Euler's function. This is a condition that guarantees that every group of order $[G:H]$ is cyclic, and includes numbers, such as $255$, that are products of more than two primes.

Solution 3:

Proposition Let $G$ be a finite group, $H$ a subgroup with $|G:H|=p$, a prime. If gcd$(|H|,p-1)=1$, or $p$ is the smallest prime dividing $|G|$, then $H$ is normal and hence $G/H \cong C_p$.

Solution 4:

Whenever $\frac{|G|}{|H|}$ has order $p$ with $p$ prime, or has order $pq$ with primes $p<q$ and $p\nmid q-1$, then $G/H$ is cyclic.

If $H$ is special, then $G/H$ is cyclic, e.g., if $H=G$.

Solution 5:

A sufficient condition (although not a necessary one) is that $\frac{|G|}{|H|}$ is a prime number.