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**aman_cc** Thanks a lot ! I think I got it. Let me write it and you can plz check.

I understand - "Note also that (i ==> (ii) ==> (iii)" In an abelian G every sub-group is normal.

Right, so here is my approach

1. Prime Factorization of d - Let $\displaystyle d = p_1^{a_1}p_2^{a_2}.....p_k^{a_k}$

2. So we have sub-gorups $\displaystyle H_1, H_2....H_k$ of order $\displaystyle p_1^{a_1},p_2^{a_2},.....,p_k^{a_k}$ respectively

3. Claim $\displaystyle H = H_1H_2....H_k$ is the desired sub-group of order = d

a. That it is a subgorup is obvious from your point (iii)

b. To prove $\displaystyle |H| = p_1^{a_1}p_2^{a_2}.....p_k^{a_k}$, I will use induction.

Basic logic of induction

Consider $\displaystyle H=H_1H_2$

$\displaystyle H_1\cap H_2$ is a sub-group of both $\displaystyle H_1$ and $\displaystyle H_2$. But $\displaystyle gcd(H_1,H_2)=1$, thus $\displaystyle |H_1\cap H_2|$ =1

Hence the result by your point# (iv)

I think I have got it. If you could plz let me know if I missed/mistaked somewhere.

And, thanks for your help.