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Let $G=\{1,2,3,4,5,6\}$ and we define an operation $*$ on $G$ by $a*b=ab$, the remainder of $ab\bmod 7$.

For instance, $$2*4=8=1$$ $$5*6=30=2$$

I want to show that $(G, *)$ is a group.

I know how to show that is has a binary operation, so every result of the operation is inside $G$.

I also know how to prove the inverse. However, how to show that it is associative and has an identity element?

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    For the identity, you need $a*e=e*a=a$ for all $a$ in $G$. Can you think of which element of $G$ would satisfy that requirement? – 2017-02-07
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    would the identity element be just 1? – 2017-02-07
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    Right, very good. – 2017-02-07
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    For the associative property, you can use that ($\mathbb{Z}_7, \cdot)$ is a group in itself, so $(a \cdot b) \cdot c = a \cdot (b \cdot c)$, ($mod \ 7$) – 2017-02-07
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    @chelivery what are you talking about? The element $0\in\Bbb Z_7$ doesn't have an inverse, so it's not a group under multiplication – 2017-02-07
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    Chelivery : firstly it's $(\mathbb{Z}_7 ^*, \cdot)$ that's a group, and secondly, do you think that if OP knew that it was, he would be asking this question ? He obviously has just begun studying groups, and so it's a good exercise trying to prove precisely what you're telling him to use – 2017-02-07
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    Alex: you're right, I meant without the $0$ as Max corrected (I was thinking of the field). Also I guessed he already knew some examples of groups that he could use such as those. – 2017-02-07
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    I'm curious how you are happy to prove the inverse without knowing the identity. – 2017-02-07

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This answer was incorrect, see the answer below

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    This answer is not correct. The group he describes is not what you say it is and it is definitely not a field. – 2017-02-07
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    It should form a field if we include the typical notion of addition, right? I made that change. – 2017-02-07
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    @TylerKharazi It needs the additive identity as well – 2017-02-07
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    Absolutely not right. It is simple enough to check the axioms of a field, and check that what OP describes does not satisfy all of them. You should remove your answer. – 2017-02-07
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    You're right. I can't delete the answer if it has been accepted though. I can edit out the field bit. – 2017-02-08
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To prove associativity, consider the fact that you are multiplying integers together and then reducing modulo 7 (in your words - finding the remainder).

The remainder will not change if you change the order of multiplication, because multiplication of integers is associative.

I believe your problem with the identity is solved in the comments.

EDIT: Your group is NOT a field. $\mathbb{Z}_7$ is different to the group which you describe.