why composition series for groups are of finite length? def : a composition series of a group G is a subnormal series of finite length
$$ 1 \vartriangleleft H_1 \vartriangleleft H_2 \vartriangleleft \cdot   \cdot\cdot\cdot\cdot\cdot\cdot \vartriangleleft H_n = G $$
with strict inclusions, such that each $H_i$ is a maximal strict normal subgroup of $H_{i+1}$ .
question : can there be an subnormal series of infinite length with strict inclusions such that for all $i$ , $H_i$ is a maximal strict normal subgroup of $H_{i+1}$.Is this even true for uncountable indexed chain of subgroups.Why the finiteness is important in the def ?
 A: One reason for finiteness is that in all composition series we absolutely want Jordan-Hölder to hold.
That may become a bit problematic if we allow infinite series. If we look at countable series $$1\lhd \cdots \lhd H_{n}\lhd H_{n-1}\lhd\cdots\lhd H_1\lhd G$$
a natural "terminating condition" is to insists that
$$
\bigcap_{i=1}^\infty H_i=\{1\}.
$$
But if only require this we lose Jordan-Hölder. This can be seen already with the infinite cyclic group $\Bbb{Z}$.
We get one sequence of maximal normal subgroups by declaring $H_n=2^n\Bbb{Z}$. In this case all the composition factors are $H_n/H_{n+1}\simeq C_2$. On the other hand, if $2=p_1<p_2<p_3<\cdots$ are all the prime numbers listed in sequence, and we declare
$$H'_k=(p_1p_2\cdots p_k)\Bbb{Z},$$
then we get a different set of composition factors as this time
$$
H'_{k-1}/H'_{k}\simeq C_{p_k}.
$$
Observe that, by the fundamental theorem of arithmetic, both "normal series" here satisfy the terminating condition
$$
\bigcap_{i=1}^\infty H_i=\{0\}=\bigcap_{i=1}^\infty H'_i,
$$
and in both cases all the quotients were simple.

I'm sure more can be said, but I am not the right person to try. Leaving this as an example and/or the first potential problem that occurred to me.
