0
$\begingroup$

I'm preparing for my math exam on Tuesday and I'm currently doing old exams, which unfortunately have no solutions online, so I'm probably gonna be asking a lot of question on here the next couple of days.

Anyway, regarding this question, I'm unsure where to start with it, this is the second part of a question where the first part was, if K is a field and a, b $\epsilon \hspace{0,1 cm} K $then ab = 0 means that a = 0 or b = 0, and then I had to show if that's valid or not in case K was a ring. I just gave a counter example with $\mathbb{Z}_6$ where I argued that 2 * 3 = 0.

I don't see how that relates to the second part of the question which goes like this:

Let H be a group with p $\epsilon \hspace{0,1 cm} \mathbb{P}$ elements. Determine all sub groups of H.

Anyhelp would be appreciated.

Also, if I have multiple questions about different themes, should I post each in it's own question or just one thread for all?

Best Regards,

Rudy

  • 1
    You mean $p$ is prime? But prime order groups have no non-trivial subgroups. (after all, the order of a subgroup divides the order of the group). – 2017-02-11
  • 0
    Your hypothesis has no sense ! (i.e. let $H$ be a group with $p\in\mathbb P$, then determine all sub-group of $H$... what do you want to say ?). Do you mean that $H$ is a group of order $p$ with $p$ prime ? – 2017-02-11
  • 0
    Surb, "Let H be a group with p $\epsilon \hspace{0,1 cm} \mathbb{P}$ elements. Determine all sub groups of H." That is the actual question I'm given, it's not in English, so I had to translate it... It's basically saying, if H is a group with prime numbers as Elements, what are all of the sub groups? – 2017-02-11
  • 0
    I don't know what it means to speak of a prime number as an element of a group. I would guess that the question meant "suppose the order of a group $H$ is prime, find all subgroups of $H$". In that case, the answer would be $e$ and $H$. – 2017-02-11
  • 0
    Actually now looking at it again, I think you're right. I just literally translated the questions. I think the question is talking about the order of the group, i.e. the number of elements in the group being prime. Ok so you're saying it's gonna be "e", because the order of a subgroup has to divide the order of the Group itself, and if the order of the original group is prime, then nothing divides a prime, except 1 and the number it self, so then we would only have trivial sub groups as in H it self and the group containing {1}? – 2017-02-11
  • 0
    Yes, that's my reading of the problem. And I agree with your reasoning. – 2017-02-11
  • 0
    Yeah the question was simply poorly formed, they should have just provided the order directly, but I guess part of the test is to see if I know what the order of a group means. Thank you! – 2017-02-11

1 Answers 1

0

Your counterezample for zero-divisors is correct. Also a set of all functions $f:A\to\Bbb R$ is a ring with natural addition and multiplication. It is not difficult to find manu divisors of zero. It is enough to have $A$ as a set with at least two elements and take two disjoint subsets of $A$, say $B$ and $C$. Put $f(x)=0$ on $B$ and $f(x)=1$ on $C$ and $g$ in a similar way by interchanging the roles of $B,C$.

The field is, in particular, an integral domain. If an element is invertible, it is not a divisor of zero. Indeed, if $a$ is invertible and $ab=0$ for some $b$, then $a^{-1}ab=0$, hence $b=0$. Because all non-zero elements are invertible, a field is an integral domain.