I have a question regarding the groups of order $|G|=100=2^2\cdot 5^2$. Using Sylow's theorem we can see that ($n_p$ being the number of $p$-Sylow-Subgroups $P_p$)
- Unless I miscalculated we have $n_5=1\implies P_5\trianglelefteq G$
- Also we have $P_5\cong \mathbb Z_{25}$ since there is only one group of order $25$.
- We have either $P_2\cong V_4$ or $P_2\cong \mathbb Z_4$.
Since $|P_2|=4$ and $|P_5| =25$ are coprime we have that $P_2\cap P_5 =1$ and thus $G\cong P_2\ltimes P_5$. This means that in total I get 2 groups of order $100$. Obviously this is false.
Question: Where is my mistake?
Edit: First mistake is that there are two possibilites for $P_5$: $\mathbb Z_{25}$ and $\mathbb Z_5^2$. So in total I now get 4 groups of order $100$ which still is wrong.