Firstly, What is identity element for this group?

Question

$A = \{0,1,2,3,4,5,6, \ldots 23\}$

$a*b =(a+b)\mod{24}$

How many proper subgroups does the group $G(A,*)$ have?

Answer 1

It is cyclic group identity of this group is $0$.

According to Lagrange's theorem, the order of the subgroup divides the order of the group.

In order to find the number of proper subgroups, we need to find the proper divisors of $24$

$24 = 2^3 \times 3^1$

No of proper divisors $= 4 \times 2 - 2 = 6$

So $6$ subgroups are there.

Comments

Link to find no. of divisors

https://www.math.upenn.edu/~deturck/m170/wk2/numdivisors.html

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how the no. of proper divisor is 6?  it should be 8

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8 are total subgroups but you have to remove { },{0,1,2,.....23} these two subgroups to make proper subgroups.

6 will be answer