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Let $p\ge5 $ be a prime. Then

  1. $\mathbb{F_p\times\mathbb{F_p}}$ has atleast five subgroups of order p.

  2. Every subgroup of $\mathbb{F_p\times\mathbb{F_p}}$ is of the form ${H_1\times{H_2}}$ where ${H_1 , {H_2}}$ are subgroups of $\mathbb{F_p}$.

  3. Every subgroup of $\mathbb{F_p\times\mathbb{F_p}}$ is an ideal of the ring $\mathbb{F_p\times\mathbb{F_p}}$

  4. The ring $\mathbb{F_p\times\mathbb{F_p}}$ is a field.

1...true .Let p=5 then subgroup generated by (1,0) , (2,0) , (3,0) , (4,0) , (0,1),....are all subgroups of order 5.

2...not true consider subgroup generated by (x,x) which cannot be written as ${H_1\times{H_2}}$.

3...I am not able to conclude for this option

I know that ideals of ring corresponds to normal subgroups of corresponding group . $\mathbb{F_p\times\mathbb{F_p}}$ is abelian group so all subgroups are normal so can i say that all subgroups are normal subgroups and hence ideal ?

4...not true because in Fp×Fp (1,0)(0.1)=(0,0) hence it is not integral domain hence it is not field.

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dipali mali
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  • How about your answer for (2), can it be an ideal? – Anurag A Jun 14 '17 at 05:01
  • In 1, the subgroup generated by $(2,0)$ is the same as that generated by $(1,0)$, so alas your argument fails. – Angina Seng Jun 14 '17 at 05:02
  • As a more general exercise for 3), you can try to characterize, up to isomorphism, all the rings that have the property "All additive subgroups are ideal". This is an interesting exercise and it answers your 3rd question – Maxime Ramzi Jun 14 '17 at 05:49
  • Got it ,subgroup generated by (x,x) is not an ideal. In 1, question is not asked for distinct subgroups so can i say first option is correct? – dipali mali Jun 14 '17 at 06:50

1 Answers1

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(1) is true. Consider any element $x(k,n) = (nk\pmod p) \times k$ with $0\leq n<p$ and $k\neq 0\pmod p) $. It is easy to see that $x(k,n)$ gnerates a cyclic subgroup of $\Bbb{Z}_p \times \Bbb{Z}_p$. It is also easy to show that the group generated by $x(k,m)$ does not contain $x(k,n)$. Thus there are $p$ distinct subgroups of this form.

(2) is false. Consider the subgroup $\{ k \times k\}$ with $0\leq k<p$. This is not of the form $H_1 \times H_2$

jasmine
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