Do Gaussian integers form a group under multiplication

Could you please show a full general proof checking for identity, closed, inverse and so on
PLEASE HELP AND SHOW WORKING !!!! :)

to make thing clear, i will assume in the following:

z = a+bi, where a,b are integers
w = c+di, where c,d are integers
u = h+ki, where h,k are integers

strictly speaking, the answer straight-away is no, only the non-zero numbers could possibly qualify, since 0 = 0+0i can not possibly have a an inverse. we will see later, that other elements do not, as well.

first, we check closure:

zw = (a+bi)(c+di) = a(c+di) + (bi)(c+di) = ac + adi + bci + bd(i^2)

= (ac-bd) + (ad+bc)i, wich is in Z[i], since ac-bd and ad+bc are both integers.

next, we check associativity:

(zw)u = [(a+bi)(c+di)](h+ki)

= [(ac-bd) + (ad+bc)i](h+ki)

= (ac-bd)h - (ad+bc)k + [(ac-bd)k + (ad+bc)h]i

= ach - bdh - adk + bck + (ack - bdk + adh + bch)i

= a(ch-dk) - b(ck+dh) + (a(ck+dh) + b(ch-dk))i

= (a+bi)[(ch-dk) + (ck+dh)i] = (a+bi)[(c+di)(h+ki)]

= z(wu). associativity is proved.

let 1 = 1+0i:

then z(1) = (a+bi)(1+0i) = a(1) - b(0) + (a(0) + b(1))i

= a+bi, while (1)z = (1+0i)(a+bi) = (1)a - (0)b + ((1)b + (0)a)i

= a+bi, so 1 = 1+0i is an identity.

unfortunately Z[i] does not contain all inverses.

let v = v+0i, where v is an integer greater than 1.

suppose v had an inverse 1/v = (m+ni).

then (v+0i)(m+ni) = 1+0i.

but (v+0i)(m+ni) = vm - 0n + (vn + 0m)i = vm + vni

thus vm = 1, and vn = 0.

since v ≠ 0, this means n = 0, so 1/v is of the form m = m+0i.

but vm = 1, so m must be 1/v.

since v is an integer > 1, 1/v is a positive integer less than 1,

but no such thing exists.

No.
Obviously they're closed under multiplication, and there is an identity element 1.
But there are no inverses. Consider the number z=3+4i, for instance. We have |z|=5, so if an inverse element z* exists then it must satisfy |z|=1/5, which is impossible since Gaussian integers have integer moduli.