Prove or disprove that $H = \{n \in \mathbb{Z}\,|\, n\, \text{is divisible by $8$ and $10$}\}$ is a subgroup of $\mathbb{Z}$. Is my reasoning correct: since the integers divisible by $8$ and $10$ are only $1$ and $2$, can I say that $H=\{n \in \mathbb{Z}\,|\, n = 1, 2\}$. Now if $H$ is going to be a subgroup then it must be closed under the same binary operation as $\mathbb{Z}$, that is addition. But $1+2 = 3$, which is an element of $\mathbb{Z}$, but not an element of $H$. Hence $H$ is not a subgroup.
 A: Notice that
$$H=8\Bbb Z\cap 10\Bbb Z$$
so $H$ is a subgroup of $\Bbb Z$ (namely $\operatorname{lcm}(8,10)\Bbb Z$) as it's intersection of two subgroups.
A: As said in the comments, $1$ and $2$ are not divisible by neither
of $8$ or $10$. an integer $a$ is said to divide an integer $b$
if there is some $k\in\mathbb{Z}$ s.t $ak=b$.
The set of numbers that are divisible by $8$ are all the numbers
of the form $\{8k|\, k\in\mathbb{Z}$} and similarly for the set
of number divisible by $10$.
Hint $1$: $\langle8\rangle$ - The cyclic group generated
by $8$ is exactly the set of numbers divisible by $8$
Hint $2$$:$ The intersection of subgroups is also a subgroup 
A: I think you misunderstood the definition of $H$.  A number $n$ is divisible by $8$ if $8$ divides $n$.  So the integers that are divisible by $8$ and $10$ are integers where $8$ and $10$ both divide them.
For example, $80$ is an integer that is divisible by both $8$ and $10$.

Now, to show that $H$ is a subgroup of $\mathbb{Z}$, you can use the subgroup test.  First, show $H$ is closed under addition, and then show that if $a, b \in H$, then $a - b \in H$.  
Here is a hint:  $8\mathbb{Z} = \{ 8n \mid n \in \mathbb{Z} \}$ and $n\mathbb{Z} = \{ 10n \mid n \in \mathbb{Z} \}$ are both subgroups of $\mathbb{Z}$.  $8\mathbb{Z}$ represents the integers that are divisible by $8$, and $10\mathbb{Z}$ are the integers that are divisible by $10$.  So, the set of integers that are divisible by both $8$ and $10$ are exactly $8\mathbb{Z} \cap 10\mathbb{Z}$.
So, $H = 8\mathbb{Z} \cap 10\mathbb{Z}$. 
Guideline for proof: to prove $H$ is a subgroup using the subgroup test, you need to show it is closed.  Use the fact that any two elements that are in $H$ are in $8\mathbb{Z}$ (which is a subgroup) and also in $10\mathbb{Z}$ (which is also a subgroup).
Similarly, if $a, b \in H$, then $a, b \in 8\mathbb{Z}$, which is a subgroup, so what can be said about $a - b$?  Similarly, $a, b \in 10\mathbb{Z}$, which is a subgroup, so what can be said about $a - b$?  What does that mean for $a- b$ with respect to $H$?
