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I have a group over $\mathbb Z$, defined by the binary operation $*$, such that $a*b:=a+b+2$.

From the previous exercise, I have deduced that the identity-element is $-2$ and that it is an abelian group under $\mathbb Z$, but the next exercise wants me to prove that it is isomorphic to the cyclic group $(\mathbb Z,+)$. But I dont really know how to write the answer. In the cyclic group $(\mathbb Z,+)$, you have that $(a,b)=a+b$, and in $(\mathbb Z,*)$ you have that $(a,b)=a+b+2$.

From my understaning, I must set up an homomorphism between those two groups $(\mathbb Z,+)$ which shows to be surjective and injective, but I'm a bit stuck.

Help would be appreciated.

Martin Sleziak
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user134489
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    See http://math.stackexchange.com/questions/373731/showing-that-g-is-a-group-under-an-alternative-operation. – lhf May 22 '14 at 10:33

2 Answers2

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Hint: a homomorphism of a cyclic group is completely determined by its effect on a generator. The generators of the additive group $\mathbb Z$ are $\pm 1$.

You have identified the identity element of your group. Can you find a generator?

Also: You will probably do best to use different letters for the elements of the two groups. Using $a,b$ for both is almost certain to lead to confusion.

Mark Bennet
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You have discovered that the original identity element $0$ has moved 2 units to the left on the number line. How does this suggest moving other elements?

Lee Mosher
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