Let's list ALL cyclic subgroups of $\Bbb Z_{10}$ (remember we will work mod 10). It starts out pretty easy:
$\langle 0\rangle = \{0\}$
$\langle 1\rangle = \{0,1,2,3,4,5,6,7,8,9\} = \Bbb Z_{10}$
$\langle 2 \rangle = \{0,2,4,6,8\}$
With 3, it gets a little tricky: we start out
$\langle 3\rangle = \{0,3,6,9,\dots\}$
and then something unusual happens:
9+3 = 12 = 2 (mod 10).
so if we continue adding 3's, we get:
$\langle 3\rangle = \{0,3,6,9,2,5,8,1,4,7\}$
so this is ALSO all of $\Bbb Z_{10}$, just "in a different order".
So $\langle 3\rangle = \Bbb Z_{10}$ (3 is a generator just like 1 is).
OK, let's move on to 4. We have:
$\langle 4\rangle = \{0,4,8,2,6\}$. Note that this is the same subgroup as $\langle 2\rangle$.
5 is pretty-straightforward, we have:
$\langle 5\rangle = \{0,5\}$ (5 is of order 2 in $\Bbb Z_{10}$ since 5+5 = 0, the identity).
Ok, now 6:
$\langle 6\rangle = \{0,6,2,8,4\} = \langle 2\rangle = \langle 4\rangle$.
And 7:
$\langle 7\rangle = \{0,7,4,1,8,5,2,9,6,3\}$ (I have listed the sums in order:
7+7 = 14 = 4 (mod 10)
7+7+7 = 14+7 = 4+7 = 11 = 1 (mod 10), etc.
just so you understand what is going on).
Just two more possible subgroups to go:
$\langle 8\rangle = \{0,8,6,4,2\}$
$\langle 9\rangle = \{0,9,8,7,6,5,4,3,2,1\}$
As you can see we only get 4 DISTINCT subgroups:
$\Bbb Z_{10} = \langle 1\rangle = \langle 3\rangle = \langle 7\rangle = \langle 9\rangle$ of order 10,
$\langle 2\rangle = \langle 4\rangle = \langle 6\rangle = \langle 8\rangle$ of order 5,
$\langle 5\rangle$, of order 2,
$\langle 0\rangle$ of order 1.
In fact, there is a nifty formula for determining the size of $\langle k\rangle$ in $\Bbb Z_{10}$, it is:
$|\langle k\rangle| = \dfrac{10}{\text{gcd}(k,10)}$
which you may want to verify for yourself using the discussion above.
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Now, what do we MEAN when we say $\Bbb Z_{10} = \langle 2,5\rangle$?
We mean any element of $\Bbb Z_{10}$ can be written as:
$k2 + m5$, for some INTEGERS $k,m$. Note that $k2$ is this context does not mean:
"$k$ times 2" as an integer, but rather:
$2 + 2 +\cdots + 2\ (k\text{ times})$ mod 10.
Furthermore we take:
$02 = 0$ and
$(-k)2 = k8$, if $k > 0$ (since "negative 2" that is, the additive inverse of 2, is 8 mod 10).
It should be obvious that the elements of $\langle 2\rangle = \{0,2,4,6,8\}$ can be written this way (use m = 0). Clearly taking k = 0, m= 1, we also get 5, so we just need to produce the elements 1,3,7 and 9 in such a fashion. The proof is in the pudding:
(3)2 + 5 = (2 + 2 + 2) + 5 = 6 + 5 = 1 (mod 10)
(9)2 + (3)5 = 8 + 5 = 3 (mod 10) (we could also use k = 4, and m = 1).
(21)2 + (7)5 = 2 + 5 = 7 (mod 10) (we could also use k = 1, and m = 1).
(27)2 + (9)5 = 4 + 5 = 9 (mod 10) (we could also use k = 2, and m = 1).
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The more interesting fact, is if we define the product in $\Bbb Z_2 \times \Bbb Z_5$ as:
$(a,b) + (a',b') = (a+a'\text{ (mod }2),b+b'\text{ (mod }5))$,
we obtain a group with 10 elements isomorphic to $\Bbb Z_{10}$ with generator $(1,1)$. A similar observation for any co-prime integers $m,n$ underlies what is known as the Chinese Remainder Theorem.
