Abstract algebra (isomorphism questions)

Let G be a group of order 4. Then
(1)If G is not cyclic, then prove that a^2=e for all a∈G and G ≃ Z_2 × Z_2.
(2)Prove that any group of order 4 is isomorphic to either Z_4 or Z_2 × Z_2.
(3)Prove that U_8 ≃ Z_2 × Z_2.
(4)Prove that U_10 ≃ Z_4.
(5)Prove that U_12 ≃ Z_2 × Z_2.
Note: Z_2 is subscript
U_8={ [a]|(a,8)=1 }={[1],[3],[5],[7]}

(1) By Lagrange's Theorem, any element of G has order dividing |G| = 4. Since G is not cyclic, any nontrivial element g ∈ G can not have order 4. Hence, g has order 2.

Since e^2 = e as well, we have a^2 = e for all a ∈ G.

Let a and b be distinct nontrivial elements of G (giving us a total of three elements of G). By closure axioms, we must have ab = ba as the fourth and final element of G.
(That ab = ba follows directly from (ab)^2 = (ab)(ab) = abab = e, and multiplying both sides on the left by a and on the right by b.)

Then, an isomorphism f : G → Z₂ x Z₂ is explicitly given by f(e) = (0, 0), f(a) = (1, 0), f(b) = (0, 1), and f(ab) = (1, 1). (The main thing to check is that f is a homomorphism, which is straightforward.)

(2) By (1), if G is not cyclic, then G ≃ Z₂ × Z₂. Otherwise, G is cyclic, and thus G ≃ Z₄.

(3) U₈ denotes the multiplicatively invertible elements mod 8 = {1, 3, 5, 7}, a group of order 4. Since 1 has order 1 and 3, 5, 7 all have order 2, we conclude by (1) that U₈ ≃ Z₂ × Z₂.

(4) U₁₀ denotes the multiplicatively invertible elements mod 10 = {1, 3, 7, 9}, a group of order 4. Since 3 has order 4 (as 3^2 = -1 mod 10, and 3^4 = 1 mod 10), we conclude that by (1) that U₁₀ ≃ Z₄ (that is, U₁₀ is cyclic).

(5) This is like (3).
U₁₂ = {1, 5, 7, 11}, a group of order 4. Since 1 has order 1 while 5, 7, 11 all have order 2, we conclude by (1) that U₁₂ ≃ Z₂ × Z₂.

I hope this helps!

1) Order O(g) of any element g divides order of the group G. Thus O(g) divides 4 and equals 1, 2 or 4. But G is not cyclic so O(g) is not equal 4. Then O(g) = 1 for neutral element or O(g) = 2 for any other element. Then g^2 = e for any element by definition of the element's order. Any bijective map f between G and Z_2 \times Z_2 such that f(e) = e is an isomorphism.

2) If G is cyclic, then G = Z_4. If not, see (1) above

3) U_8 is not cyclic (O(1) = 1, O(3) = O(5) = O(7) = 2). Thus U_8 = Z_2 \times Z_2 (see (2))

4) U_10 = {1, 3, 7, 9} is cyclic (O(3) = 4) thus U_10 = Z_4

5) U_12 = {1,5,7,11} is not cyclic (O(5) = O(7) = O(11) = 2) thus (see (1)) U_12 = Z_2 \times Z_2

"\times" means "×"
G = H means G ≃ H