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As I said before in my previous posts i am learning abstract algebra from Herstein's book in order to enhance my knowledge in that area. Quite often I will upload problems on that branch of math and will be very thankful if you can help me or check out my proof.

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My solution:

(a) Let $G=\mathbb{Z}, \ a\cdot b=a-b$. We see than binary operation holds in that system. However, associativity fails since $(a-b)-c\neq a-(b-c)$. Also there is no identity element i.e. $e$ in that system and the property of existense of inverse element is pointless since we have no $e$. $G$ is not a group.

(b) Let $G=\mathbb{N}$ with $\ a\cdot b=ab$. It's easy to check that binary operation, associativity and the existence of identity element hold in $G$. However, the property of inverse element fails. For instance, $2\in G$ but there is no element $a$ from $G$ such that $2a=1$. $G$ is not a group!

(d) It's easy to verify that binary operation, associativity, identity element ($e=0$) and also inverse element holds in system $G$. Thus $G$ is a group.

(c) It is the most interesting part of that problem, especially associativity. It's easy to see that $a_i\cdot a_j=a_{i+j \pmod 7}$. Binary holds in this system since $i+j \pmod 7 \in \{0,1,2,3,4,5,6\}$. Identity element also exists, namely $e=a_0\in G$. Inverse element also exists, since $(a_0)^{-1}=a_0$ and for $i: 1\leqslant i\leqslant 6$ we have $(a_i)^{-1}=a_{7-i}\in G$. In order to check associativity we should check the following identity: $(a_i\cdot a_j)\cdot a_k=a_i\cdot (a_j\cdot a_k)$. The left-hand side is equal to $$a_{i+j \pmod7} \cdot a_k=a_{((i+j)\pmod 7+k)\pmod 7}$$ And the right-hand side is equal $$a_i\cdot a_{j+k \pmod7}=a_{(i+(j+k)\pmod 7)\pmod 7}$$ However, using the elementary number theory it is easy to check that $((i+j)\pmod 7+k)\pmod 7=(i+(j+k)\pmod 7)\pmod 7$. Thus $G$ is a group.

Is my reasoning true?

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Your answers are correct but a few of your proofs could be improved a bit.
For (a) you should show an example of non-associativity.
For (d) you need to show that the group operation is well-defined even though a given rational number has different representations.

Generally, specific examples should be shown for the identity, inverses, and counterexamples to universal statements.
If you were handing these problems as homework, you wouldn't normally get away with "It's easy to verify ..." You might get away with doing something once and then referring to it.
On the other hand if this were a mathematical paper then you might not say anything except that it's a group unless verifying one of the group axioms is tricky.

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  • $\begingroup$ Is there a problem with my answer? If not, please accept it. Thanks. $\endgroup$ – Stephen Meskin Nov 22 '17 at 8:36

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