So I am currently trying to Classify all groups of order 100 through an extensive proof; and this is as far as I have gotten so far, wondering how to go beyond the fact that both squares (Z4 & Z25) are isomorphic to the group, any help would be great.
Classify all groups of order 100.
G=|100|
P=C25P=C25, Q=C4Q=C4, |im(ϕ)|=1|im(ϕ)|=1: 11 group;
P=C25P=C25 , Q=C4Q=C4 , |im(ϕ)|=2|im(ϕ)|=2 : 11 group;
P=C25P=C25 , Q=C4Q=C4 , |im(ϕ)|=4|im(ϕ)|=4 : 11 group;
P=C25P=C25 , Q=C22Q=C22 , |im(ϕ)|=1|im(ϕ)|=1 : 11 group;
P=C25P=C25 , Q=C22Q=C22 , |im(ϕ)|=2|im(ϕ)|=2 : 11 group;
P=C25P=C25 , Q=C22Q=C22 , |im(ϕ)|=4|im(ϕ)|=4 : not possible;
P=C25P=C52 , Q=C4Q=C4 , |im(ϕ)|=1|im(ϕ)|=1 : 11 group;
P=C25P=C52 , Q=C4Q=C4 , |im(ϕ)|=2|im(ϕ)|=2 : 22 groups;
P=C25P=C52 , Q=C4Q=C4 , |im(ϕ)|=4|im(ϕ)|=4 : 44 groups;
P=C25P=C52 , Q=C22Q=C22 , |im(ϕ)|=1|im(ϕ)|=1 : 11 group;
P=C25P=C52 , Q=C22Q=C22 , |im(ϕ)|=2|im(ϕ)|=2 : 22 groups;
P=C25P=C52 , Q=C22Q=C22 , |im(ϕ)|=4|im(ϕ)|=4 : 11 group.
These numbers are prime, lagrange implies prime groups are cyclic therefore cyclic groups, only one group of each order (up to isomorphism).
Z4
If G has an element of order 4, it is cyclic, assume no element of order 4 G={e,a,b,c}
order a = order b = order c = 2
claim ab = c
if ab = e, then a = b^-1, contradicts b = b^-1
ab = a implies b = e
while ab = b implies a = e
same argument shows ba = c = ab, ca = b = ac, cb = a = bc
G: Z2 x Z2 is isomorphic to G
f(e) = (0,0) f(a) = (1,0), f(b) = (0,1), f(c) = (1,1)
Tips on moving forward?
If $|G|=100$ then $G = P \rtimes_\phi Q$ with $P\in{\rm Syl}_5(G)$ and $Q \in {\rm Syl}_2(G)$. Here is a summary of the number of isomorphism classes of groups in the various situations. Most of these are not hard to prove.
$P=C_{25}$, $Q=C_4$, $|{\rm im}(\phi)|=1$: $1$ group;
$P=C_{25}$, $Q=C_4$, $|{\rm im}(\phi)|=2$: $1$ group;
$P=C_{25}$, $Q=C_4$, $|{\rm im}(\phi)|=4$: $1$ group;
$P=C_{25}$, $Q=C_2^2$, $|{\rm im}(\phi)|=1$: $1$ group;
$P=C_{25}$, $Q=C_2^2$, $|{\rm im}(\phi)|=2$: $1$ group;
$P=C_{25}$, $Q=C_2^2$, $|{\rm im}(\phi)|=4$: not possible;
$P=C_{5}^2$, $Q=C_4$, $|{\rm im}(\phi)|=1$: $1$ group;
$P=C_{5}^2$, $Q=C_4$, $|{\rm im}(\phi)|=2$: $2$ groups;
$P=C_{5}^2$, $Q=C_4$, $|{\rm im}(\phi)|=4$: $4$ groups;
$P=C_{5}^2$, $Q=C_2^2$, $|{\rm im}(\phi)|=1$: $1$ group;
$P=C_{5}^2$, $Q=C_2^2$, $|{\rm im}(\phi)|=2$: $2$ groups;
$P=C_{5}^2$, $Q=C_2^2$, $|{\rm im}(\phi)|=4$: $1$ group.
The hardest case is probably $P=C_{5}^2$, $Q=C_4$, $|{\rm im}(\phi)|=4$. There is one group in which ${\rm im}(\phi)$ acts trivially on one direct factor of $P$, one in which it acts as a group of order $4$ on one factor and $2$ on the other factor, and two groups in which it acts as groups of order $4$ on both factors.
It might help to think of ${\rm im}(\phi)$ as a subgroup of ${\rm GL}(2,5)$.