Group containing both $5\mathbb{Z}$ and $7\mathbb{Z}$ Prove that there doesnot exist any proper subgroup of $(\mathbb{Z},+)$ containing both $5\mathbb{Z}$ and $7\mathbb{Z}$ 
 Since $5$ and $7$ are coprime hence there cannot exist any proper subgroup containing multiples of both $5$ and $7$ if it contains then it must be  the group $\mathbb{Z}$ itself 
Is m above reason correct please help?
 A: You've basically just rephrased the problem, throwing in the word coprime, which is essential, but you haven't really shown how. Unless you have a theorem somewhere saying something like "No proper subgroup of $(\Bbb Z,+)$ can contain two coprime elements", I wouldn't find that enough.
Try this: show specifically why a subgroup containing both $5$ and $7$ must contain $1$.
A: If a subgroup $G \le (\mathbb{Z}, +)$ contains $5\mathbb{Z}$ and $7\mathbb{Z}$ then it also contains
$$1 = 50 - 49 = 10\cdot 5 - 7 \cdot 7$$
so $G = \mathbb{Z}$.
A: The key is that 5 and 7 are indeed coprime, but the numbers themselves are irrelevant. In fact, you might see it clearer if you turn to a more general situation:
Claim. If $m,n \in \mathbb{Z}$ are two coprime integers, then the only subgroup of $(\mathbb{Z},+)$ containing $m\mathbb{Z}$ and $n\mathbb{Z}$ is precisely $\mathbb{Z}$.
Proof. Let $H \subseteq \mathbb{Z}$ be a subgrup of $(\mathbb{Z},+)$ containing $m\mathbb{Z}$ and $n\mathbb{Z}$. Clearly $\mathbb{Z}=1\mathbb{Z}$, so it suffices to show that $1 \in \mathbb{Z}$.
Since $m$ and $n$ are coprime, by Bézout's identity we have that
$$mx+ny=1$$
for some $x,y \in \mathbb{Z}$.
Besides, $m,n \in H$ and $H$ is closed under addition, so it follows that $1 \in H$ and we are done.
