# The verification of axioms of the group in certain examples

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.

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?