If #x# generates the group #G#, which has order #25#, it means that, as a set, #G# is composed by
#G = \{x, x^2, x^3, ..., x^25\}#
Now, we're supposing that #alpha# is a factor of #25#. This means that there exists #k \in \mathbb{N}# such that #k*alpha = 25#. So, the group generated by #x^alpha# is
#\hat{G} = \{x^alpha, x^{2alpha}, ... x^{kalpha}\}#.
This means that every element #x^s#, which #s \ne t alpha# for any #t#, will not be in #\hat{G}#.
After all, the only factors of #25# are #1,5# and #25# itself.
If #alpha=1#, then #x# and #x^alpha# are the same element, and thus generate the same group.
If #alpha=5#, then the group generated by #x^5# is
#\hat{G} = \{x^5, x^{10}, x^{15}, x^{20}, x^{25}\}#.
In fact, #x^30=x^5# because #x# has order #25#. So, you can see that #G# and #\hat{G}# are not the same group.
Finally, if #alpha=25#, you have the trivial group
#\hat{G} = \{x^{25}=e\}#
