#theorem32#
\ \ If a group has a composition series of length
one, then the group is simple and any two composition series are certainly
equivalent. Now suppose that a group G has a composition series
#equation34#
of length n;SPMgt;1, and that if a group has a composition series of length less
than n, then any two composition series of that group are equivalent. Let
#equation37#
be any composition series of G. Consider the series
#equation40#
and
#equation43#
Since #tex2html_wrap_inline215# is a normal subgroup of #tex2html_wrap_inline217# and #tex2html_wrap_inline219#, the Third Isomorphism Theorem (<#48#>2.3.12<#48#>) yields
#displaymath146#
#displaymath147#
and #tex2html_wrap_inline221# is a normal subgroup of #tex2html_wrap_inline223# since it is a
product of two normal subgroups. Since #tex2html_wrap_inline225# is a simple group, #tex2html_wrap_inline227# is either #tex2html_wrap_inline229# or #tex2html_wrap_inline231#. That is, #tex2html_wrap_inline233# is either #tex2html_wrap_inline235# or #tex2html_wrap_inline237#. Therefore,
if we remove repetitions from
#displaymath148#
we get a composition series for #tex2html_wrap_inline239#. By our induction hypothesis, the
resulting composition series is equivalent to the composition series
#displaymath149#
and hence (#one#63>) and (#three#64>) (with repetitions removed) are
equivalent. #tex2html_wrap_inline241#