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  1. $(\mathbb{Z},+)$ and $(\mathbb{Z}, *)$ where $a*b=a+b-1$
  2. $G$ and $G\times G$, where $G=\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2 \cdots$, one copy of $\mathbb{Z}_2$ for each positive integer.
  3. $(\mathbb{Q},+)\times (\mathbb{Q},+)$ and $(\mathbb{Q},+)\times (\mathbb{Z},+)$

Can any one help me with these three? Either hint or answer will be greatly appreciated. I just don't know what group property to look for to see whether they are isomorphic or not. Or in the first place I don't even have an intuition about whether they are isomorphic, and using a specific mapping doesn't really help. For this reason I don't even know whether I should try to prove that they are isomorphic, or I should look for some unsatisfied group property. Any comments or critics on my approach are also appreciated.

Thanks everyone.

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HINTS:

  1. Work out what the identity and inverses are in $\langle\Bbb Z,\ast\rangle$ to see just what its structure is, and try to compare it with $\langle\Bbb Z,+\rangle$.
  2. There is a bijection between $\Bbb Z$ and $\Bbb Z\times\Bbb Z$.
  3. Is it possible to solve the equation $x+x=a$ for every $a$ in the first group? What about the second?
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  • $\begingroup$ for 1.I am getting first group cyclic & the second not,is it correct? $\endgroup$ – Learnmore Oct 29 '14 at 4:01
  • $\begingroup$ @learnmore: No, the second group is cyclic. What do you have for the identity, and if $n\in\Bbb Z$, what is the inverse of $n$? $\endgroup$ – Brian M. Scott Oct 29 '14 at 4:10
  • $\begingroup$ My identity is 1 & inverse of any $a$ is $2-a$ $\endgroup$ – Learnmore Oct 29 '14 at 4:11
  • $\begingroup$ @learnmore: Good; those are right. Now what happens when you calculate $2,2*2,2*2*2,2*2*2*2,\ldots$? You get ... ? $\endgroup$ – Brian M. Scott Oct 29 '14 at 4:13
  • $\begingroup$ 2,3,4.. you are right i.e. generated by 2 $\endgroup$ – Learnmore Oct 29 '14 at 4:15

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