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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 $G$.

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    $\begingroup$ $G$ is abelian and $H$ is the center of $G$ :p $\endgroup$
    – Exit path
    Aug 3, 2018 at 18:11

7 Answers 7

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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.

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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.

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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 $G/H \cong C_p$.

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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 $H=G$.

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  • $\begingroup$ What is a special subgroup? $\endgroup$
    – lhf
    Aug 3, 2018 at 21:18
  • $\begingroup$ I wanted to say, if we chose a special subgroup, e.g., some specific example like $H=G$, as a "good condition on $H$". $\endgroup$ Aug 4, 2018 at 8:05
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A sufficient condition (although not a necessary one) is that $\frac{|G|}{|H|}$ is a prime number.

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Let $G$ be a group and $H$ be a normal subgroup, the easiest condition that I can think of is that $\frac{|G|}{|H|}$ is a prime number.

If $G$ is not a solvable group and $H$ is solvable, then $G/H$ must not be cyclic.

Another equivalent condition that you may find useful is that, $G/H$ is cyclic if and only if for any $g\in G\backslash H$, the minimal number $n$ such that $g^n\in H$ must satisfy $n=\frac{|G|}{|H|}$.

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It is of course a necessary condition that $H$ contains the derived subgruop $[G,G]$ of $G$, as $G/H$ is commutative. Suppose that you have a set $X$ of generators for $G$. You could then check that, for each prime divisor $p$ of the index $[G:H]$, the subgroup $\langle x^p\mid x\in X\rangle H$ has index $p$ in $G$. (This just amounts to saying that the $p$-th powers of a generating set for the quotient group $G/H$ generate an index $p$ subgroup.)

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