Show $\{1,3,5,7,9,11,13,15\} \cong \mathbb{Z}_2 \times \mathbb{Z}_4$ 
Show $\{1,3,5,7,9,11,13,15\} \cong \mathbb{Z}_2\times\mathbb{Z}_4$ where $G$'s binary operation is multiplication mod $16$.

I've determined the orders of all the elements: $1$ has order $1$; $3$ has order $4$; $5$ has order $4$; $7$ has order $2$; $9$ has order $2$; $11$ has order $4$; $13$ has order $4$; and $15$ has order $2$.
Since the elements have orders $1, 2$ and $4$, it seems that this is going to lead to $G \cong \mathbb{Z}_2 \times \mathbb{Z}_4$, but why? What is the relationship between the orders and $\mathbb{Z}_2 \times \mathbb{Z}_4$?
 A: This group is Abelian since multiplication is commutative. 
This group has 8 elements,
thus it can be $\mathbb Z_8$, $\mathbb Z_2 \times \mathbb Z_4$ or $\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2$.
3 has order 4.
No element has order 8.
Thus it is $\mathbb Z_2 \times \mathbb Z_4$.
A: First we need to show G = {1,3...} is a group.
1) It's close under multiplication:  All elements are odd.  the product of odd numbers are odd.  odd numbers are congruent to odd numbers modulo even numbers.  This set is a set of all odd numbers congruent modulo 16.  So its closed under multiplication.
2) An identi... oh, c'mon, you know it is "1".
3) All elements have inverses: $16 = 2^4$ so $gcd(2, \text{odd}) = 1$ so $\text{odd}^{\phi(16)} \equiv 1\mod 16$.  So $m * m^{\phi(16) - 1} \equiv 1 \mod 16$.  Yes, they all have inverses.
So G is a group.
At this point it might just be easier to one by one show each element of G is equivalent to an element of $\mathbb Z_2 \times \mathbb Z_4$.  There's only 8 of them... but theory is theory and we should do this right.
We know $G \cong \mathbb Z_2 \times \mathbb Z_4 \iff$ there is a homomorphism $f: Z_2 \times \mathbb Z_4 \rightarrow G$.
i.e. $f(a + b) \equiv f(a)*f(b) \mod 16$.
Which means we want $f(a,b) = f(a(1,0) + b(0,1)) = f(1,0)^a * f(0,1)^b$.  
So we want $f(0,1) =x$ where x has order 4.  Well, $x = 3$ has $x^4 = 81 \equiv 1 \mod 16$. (And $x, x^2, x^3 \not \equiv 1 \mod 16$).  
So define $f(0,1) = 11$.  Then $1 = f(1,1) = f(1,0)*11 \equiv 1\mod 16$ so $f(1,0) = $ ... $3$ will do it.
So we define $f(a,b) = 3^a*11^b$ should be a homomorphism.
$f((a,b) + (c,d)) = 3^{a+c}*11^{b+d} = 3^a*11^b*3^c*11^d = f(a,b)*f(c,d)$ so it is a homomorphism and
$G \cong   \mathbb Z_2 \times \mathbb Z_4 $
