H is a group with p $\epsilon \hspace{0,1 cm} \mathbb{P}$ elements (Ord(H) = p $\epsilon \hspace{0,1 cm} \mathbb{P}$) . Determine all sub groups of H.

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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

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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.