There is an element of order $51$ in the multiplicative group $(Z/103Z)^∗.$ How to show that there is an element of order $51$ in the multiplicative group $(Z/103Z)^∗.$
I can't use Cauchy's theorem since $51$ is not prime.
 A: EDIT Here's another proof in terms of Cauchy's Theorem, which the OP has learned. (My two other proofs rely on a theorem about finite fields and the structural theorem for finite abelian groups, neither of which the OP may have learned.)
Note that $103$ is prime, so $G=(\mathbb{Z}/103\mathbb{Z})^*$ has order $102=2\cdot  3 \cdot 17$. $G$ is obviously abelian. By Cauchy's theorem, $G$ contains an element $x$ of order $3$ and an element $y$ of order $17$. Let $z=xy$. What is the order of $z$? Since
$$
z^{51} = x^{51} y^{51} = (x^3)^{17} (y^{17})^3 = 1,
$$
$z$ has order dividing $51=3\cdot 17$. Since $z\ne 1$, the order of $z$ is either $3$, $17$, or $51$. Observe
$$
z^3 = x^3 y^3 = y^3 \ne 1
$$
and
$$
z^{17} = x^{17} y^{17} = x^{17} = x^2 \ne 1.
$$
Hence, the order of $z$ is neither $3$ nor $17$, so it must be $51$.

$103$ is prime, so $\mathbb{Z}/103\mathbb{Z}$ is a finite field. The multiplicative group of a finite field is cyclic. (See here for a discussion.) The cyclic group of order $102$ certainly contains an element of order $51=102/2$.
In this particular case, one can also see that $G=(\mathbb{Z}/103\mathbb{Z})^*$ is cyclic from the fundamental theorem for finitely generated abelian groups. However, note that this argument would not have worked for other choices of the prime (103 in this case). The better approach for this problem is to use the theorem quoted above.
The argument utilizing the fundamental theorem for finitely generated abelian groups goes like this:
$(\mathbb{Z}/103\mathbb{Z})^*$ is an abelian group of order $102$.  Observe $102 = 2\cdot 3\cdot 17$. By the fundamental theorem for finitely generated abelian groups,
$$
G = (\mathbb{Z}/2\mathbb{Z})\oplus(\mathbb{Z}/3\mathbb{Z})\oplus(\mathbb{Z}/17\mathbb{Z})=C_{102}.
$$
A: Here is a more general version, which does not require the group to be cyclic.
If $G$ is a finite abelian group of order $n$ and $d$ is a squarefree divisor of $n$, then $G$ has an element of order $d$.
Proof: Apply Cauchy repeatedly to get an element of order $p$ for each prime dividing $d$. Now clearly the product of these elements has order $d$ as needed (since their orders are pairwise coprime).
A: A more down-to-earth argument is by Cauchy theorem there is an element x of order 3 since 3 is a prime and divides 102 which is the order of G = (Z/103Z)*, and there is also another element y of order 17 since 17 is a prime and divides 102. Since (3,17) = 1 and G is cyclic hence abelian so the element xy then has order 3*17 = 51. 
A shorter proof is: Since G is cylic G = < x > and ord(x) = 102, so the element x^2 must have order 102/2 = 51.
Note: The claim that G is cyclic needs proof. To get an outline of proof is useful. So by Euler theorem we have for each element x in G, with x not e then x^102 = 1, and so let y = x^2, then ord(y) = 102/2 = 51, and similarly let z = x^51, then ord(z) = 102/51 = 2. (these deductions also need proof ). So since G is abelian we have the element w = yz has order 2*51  = 102 so < w >  = G since it is finite subgroup of equal size as G and this means G = < w > and therefore is cyclic. 
