Group of order three distinct primes

Question

I want to show that a group G of order 345 is Abelian.

I used Sylow's theorem to find Syl(5)=Syl(23)=1. but i was unable to conclude Syl(3)=1 because i found Syl(3)=1 or 115. I'm not sure how to proceed from here..

Please help, thank you!

score 1 · Accepted Answer · answered Dec 14 '12 at 18:09

1

Note that $(5,23)=1$ and $syl(5)=syl(23)=1$ shows that $G$ has a subgroup $H$ with order $5\times 23=115$, of course it is cyclic. And because it has index 3, which is the smallest divisor of the order of $G$, so it is normal subgroup, then we get a homomorphism $\phi$ from $C_3\to Aut(C_{115})$, while $|Aut(C_{115})|=115(1-\frac{1}{5})(1-\frac{1}{23})=88$, which is not divisable by 3, so the $\phi$ is trivial, so $G$ is a cyclic group.

answered Dec 14 '12 at 18:09

ougao

3,671

what is C3? if i tell you i haven't covered automorphism in class then how would u explain me the # 88? thanx!! – d13 Dec 14 '12 at 18:14
Why does index 3 (agreed on that) imply it is normal? – gnometorule Dec 14 '12 at 18:15
without index 3 i can say its normal because G has a normal subgroup of order 5 and 23 so their product is also normal but i just want to show that we have a normal subgroup of order 3 as well.. how do i show that if Syl(3)=115??? – d13 Dec 14 '12 at 18:22
$C_3$ is cyclic group with order 3; it is a general fact that if $H\leq G$ is a subgroup with index $p$, which is the least prime number that divides the order of $G$, then, $H$ is a normal subgroup. It is a standard exercise to prove this by using the left multiplicaton of $G$ on the left coset space of $G/H$, you can find a proof in S.Lang's "Algebra" GTM211, page 36. – ougao Dec 14 '12 at 18:24
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Proving this w/o automomorphism, note that the centralizer is not trivial (by counting the orbits of conjugacy classes, you'd get otherwise that the centralizer's orbit equals 1 mod pq for 2 of the 3 primes, which contradicts its order dividing 345). Let $H$ be the centralizer, and look at $G/H$. It will be cyclic by the above arguments, so generated by some $bH$; and $H = $ some $a \in H$. Then $a, b$ commute, = G, with only $e$ in their intersection. So $G$ is isomorphic to their product, and the 2 subgroups of the product are cyclic and so abelian, so $G$ is too. – gnometorule Dec 14 '12 at 18:35
@ougao: thanks for clarifying. Hadn't come across this. – gnometorule Dec 14 '12 at 18:38
@gnometorule welcome! – ougao Dec 14 '12 at 18:39
@gnometorule u r using the centralizer of what exactly?? – d13 Dec 20 '12 at 08:08
Centre, sorry: don't know why I typed centralizer, – gnometorule Dec 20 '12 at 13:52

calc ll · Answer 2 · 2022-08-26T17:49:20.527

Whenever $n=p_1\cdot p_2\cdots p_k$ is a product of distinct primes, and no $p_i\lvert (p_j-1)$, $n$ is a cyclic number. Groups of such order are always cyclic.

Since $345=3\cdot 5\cdot 23$, and $3\not\mid(5-1)$ and $3,5\not \mid (23-1)$, $345$ is a cyclic number.

A number $n$ is defined to be cyclic if $(n,\varphi (n))=1$. Any group of such order is cyclic, and conversely. One direction is easy, the other less so. See the references in the link. Here of course $\varphi $ is Euler's totient function.

Group of order three distinct primes

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