Is $\{(m, -m)\mid m\in \mathbb{Z}\}$ is isomorphic to $\mathbb{Z}$?
What i know about isomorphic is, simply it is a bijective homomorphism.
How one can define isomorphism map $\phi$ in above case?
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What do you mean by $(m, -m)$? Is it an interval of $\Bbb Z$ or $\Bbb R$? Is it an element of $\Bbb Z\times Z$? – Arthur Sep 07 '16 at 06:46
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@Arthur: I feel he meant the second one. – Mikasa Sep 07 '16 at 06:49
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@Arthur, Babak is right i $(m,-m)$ is a element of $\mathbb{Z} \times \mathbb{Z}$. – phy_math Sep 07 '16 at 06:56
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Yes it is ! Just take $$\varphi: \mathbb Z\longrightarrow M$$ defined by $$\varphi(m)= (m,-m).$$
The surjectivity is obvious. For the injectivity, it's easy to prove that $\ker \varphi=\{0\}.$
Surb
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The pair $(m, -m)$ is an element of $\mathbb Z^2$
$\mathbb Z$ is a set, and with the operation $+$ it is a group.
Speaking about homomorphisms between the two makes very little sensem because $(m, -m)$ is not a group (because even though technically, it is a set of two elements, you haven't defined an operation on it).
What I feel you are really trying to ask is the following question:
Is the set $\{(m, -m)| m\in\mathbb Z\}$ isomorphic to $\mathbb Z$?
In which case the answer is still unclear, because you still haven't defined an operation. Now, if the operation is $$(m, -m) + (n, -n) = (m+n, -(m+n)),$$
then yes, the two groups are isomorphic, and one possible isomorphism is the most obvious mapping you can possibly think of.
5xum
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