Let $G$ be a group with $|G|=455$. Show that $G$ is a cyclic group.

Question

Let $G$ be a group with $|G|=455$. Show that $G$ is a cyclic group.

Answer 1

It is well-known that if $n$ is a natural number, there is only one group of order n if and only if $\gcd(n,\varphi(n))=1$. Here $\varphi$ is the Euler totient function. For $n=455$ this applies. If there is only one group of a particular order it must necessarily be cyclic.

Answer 2

Hints for you to prove. Let $\,G\,$ be a group of order $\,455=5\cdot 7\cdot 13\,$ ,then:

1) There exists one unique Sylow $\,7-\,$subgroup $\,P_7\,$ , and one single Sylow $\,13-\,$ subgroup $\,P_{13}\,$ , which are then normal;

2) There exists a normal cyclic subgroup $\,Q\,$ of order $\,91\,$

3) If $\,P_5\,$ is any Sylow $\,5-\,$ subgroup, then we can form the semidirect product $\,Q\ltimes P_5\,$

4) As the only possible homomorphism from a group of order $\,91\,$ to a group of order $\,4\,$ is the trivial one, the above semidirect product is direct .

Answer 3

Let $|G|=pqr$ and WLOG $p<q<r$. Your case is the particular one for $p=5$, $q=7$ and $r=13$, where the fortunate conditions $p\nmid q-1$ and $p\nmid r-1$ and $q\nmid r-1$ take place, which allow the following counting argument.

By Sylow III, $n_p\mid qr$ and $n_p\equiv 1\pmod p$. So, $n_p=1,q,r,qr$ and $n_p=1+kp$. If $p\nmid q-1$ and $p\nmid r-1$, then $n_p\ne q,r$ and hence $n_p=1,qr$. Likewise, $n_q\mid pr$ and $n_q\equiv 1\pmod q$. So, $n_q=1,p,r,pr$ and $n_q=1+lq$. Now, $q\nmid p-1$ (because $q>p$); so, if $q\nmid r-1$, then $n_q\ne p,r$ and hence $n_q=1,pr$.

Suppose there are $qr$ $p$-Sylow subgroups and $pr$ $q$-Sylow subgroups. Since all these subgroups intesect pairwise trivially (they have order $p$ or $q$), their union's size amounts to $qr(p-1)+pr(q-1)+1$, which is greater than $pqr$$^\dagger$. Therefore, there isn't enough room in $G$ for so many $p$-Sylows and $q$-Sylows at the same time, and hence either $n_p=1$ or $n_q=1$.

If $n_p=1$, then the only $p$-Sylow is normal, say $H$. As such, $H$ is the union of conjugacy classes of $G$, with as many singletons as the order of $|H\cap Z(G)|$. But, by Lagrange, $|H\cap Z(G)|=1,p$, and the former option is ruled out because there aren't conjugacy classes (of sizes $p$, $q$, $r$ and their pairwise products) "filling the gap" $p-1$. Therefore $H$ is central, and being $G/H$ cyclic (as $q\nmid r-1$), then $G$ is abelian and finally cyclic (take the product of any three elements of order $p$, $q$ and $r$, respectively).

Same argument in case of $n_q=1$ (say $K$ the only $q$-Sylow, normal), with the only difference that now "filling non-centrally the gap" $q-1$ is prevented by the assumption $p\nmid q-1$, and the ciclicity of $G/K$ is ensured by $p\nmid r-1$.

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$^\dagger$In fact: $qr(p-1)+pr(q-1)+1=$ $2pqr-r(p+q)+1>pqr\iff$ $pqr-r(p+q)+1>0$, which is true because $pqr-r(p+q)+1>$ $pqr-2qr+1=$ $qr(p-2)+1>0$.

Answer 4

Notice $455 = 13*7*5$ and we know $13$, $7$ and $5$ are prime. Now use Lagrange theorem to show that G is cyclic. This is a hint.