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Let $p$ be a prime. Then there is a nonabelian semidirect product of $\mathbb{Z}_{p^2}$ and $\mathbb{Z}_p$. There is also a nonabelian semidirect product of $\mathbb{Z}_p\oplus\mathbb{Z}_p$ and $\mathbb{Z}_p$. Since there are only two nonabelian groups of order $p^3$, are these groups isomorphic?

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They are not. The second of your examples, but not the first, has all nontrivial elements having order p. This is because for any nontrivial element z of the acting subgroup, there is a base x,y for the normal subgroup for which the action maps x to x and y to x+y (or xy in multiplicative notation). Raising any $x^iy^jz$ to the pth power then introduces $jp(p-1)/2$ more x's when you collect terms and move z's past y's that many times. Since p is odd, this is divisible by p, and since x has order p, this means it's the same as raising everything else to the p and introducing no additional x's, which obviously gives the identity.

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