Picture the quaternion integers as a lattice in 4 dimensions. If it will help you visualize the geometry, restrict attention to 3 of the 4 dimensions. (None of us sees 4 dimensions very well.) This "slice" of the quaternions produces the set of points in 3 space with integer coordinates. Space is perfectly packed with cubes, or hypercubes, if you step back up to 4 dimensions.
Show that |x| is 0 iff x = 0. Remember that |x| is a sum of squares, hence it is positive or 0.
Then, xy = 0 means |xy| = 0, |x|×|y| = 0, |x| or |y| is 0, x or y is 0, and the quaternion integers form a domain.
This ring contains a multiplicative subgroup consisting of ±1 ±i ±j ±k. This is a nonabelian group of order 8, often denoted Q8. Note that Q8 is precisely the units of the ring, the elements with norm 1. It has a normal subgroup Z4, generated by i, and since this subgroup has index 2, it is normal in Q8. There are no involutions, other than 1, hence there is no copy of the factor group Z2 in Q8. Q8 is not a semidirect product, it is merely a nonabelian extension of Z4 by Z2.
This ring is a free Z module, spanned by 1 i j and h. But is it really a ring?
The only tricky part is multiplication. Integers times integers produce integers, so multiply a half integer quaternion by a "regular" quaternion. If a component in the latter is even, it contributes an integer to each component in the product. An odd entry, such as 13j, adds a half integer to each component in the product. At the end of the day all four components are integers or half integers.
Next, multiply two half integer quaternions together. Reduce the numerators mod 4 and watch what happens. Negate each factor if necessary, so the constant is 1 mod 4. If a factor looks like 1-i… mod 4, premultiply by i, so that the first two coefficients are 1 mod 4. The remaining coefficients, on j and k, may be 1 or -1 mod 4. That's 16 possibilities. Try them out and see. The result is always 0 mod 4 or 2 mod 4 across the board. Bring in the denominators, ½×½ = ¼, and find integers or half integers. The product does indeed belong to our ring.
If x consists of half integers, some associate of x consists of integers. Multiply by a half unit u, and if that does not do the trick, that is, ux still shows half integers, then multiply by u+1 or u-1, the "other" half unit, which adds or subtracts x to the product, which then creates integers.
To visualize the half quaternions, start with the quaternions, which describe perfect unit hypercubes in 4 space, and add their centers. Each center has half integer coefficients across the board.
Consider the norm of one of these centers, an element with half integer coefficients. The square of an odd number is 1 mod 4. Add four of these squares together to find a multiple of 4. Divide by 4 and the norm is an integer. All the norms map back into Z.
An element is a unit iff its norm is a unit, iff its norm is 1. The group of units includes the subgroup Q8, plus another 16 units, produced by negating any or all of the coefficients of h. The group of units has order 24, and is denoted Q24.
Plot the points of Q24 in 4 dimensional space. Actually, they all lie on the unit sphere. The members of Q8 are ±1 on the four coordinate axes. The other 16 points lie on the main diagonals of the 16 regions walled off by the four hyperplanes.
Multiplication by any of these units implements a linear transformation on 4 space, as described in the previous section. Therefore Q24 is a subgroup of the rigid rotations and reflections of 4 space. For instance, multiplication by -1 reflects everything through the origin, and multiplication by i rotates the xi plane and the jk plane simultaneously.
Negate h, so that h = -(1+i+j+k)/2. Then verify that h3 = 1. When the linear transformation defined by h is applied 3 times, quanternian space returns to its original orientation.
Note that h, with order 3, cannot lie in Q8. It represents a nontrivial coset of Q8, and h2 represents another coset. Finally h3 represents Q8 again.
Show that hi/h = j, hj/h = k, and hk/h = i. These generate Q8, hence the conjugate of Q8 = Q8, for h, for h2, and for all of Q24. Q8 is normal in Q24, with a factor group of Z3. Remember, a copy of Z3 lives in Q24, generated by h. Therefore Q24 is a semidirect product of Q8 by Z3. The automorphism associated with h, i.e. h*Q8/h, cycles i j and k.