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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 $P_p$)

  1. Unless I miscalculated we have $n_5=1\implies P_5\trianglelefteq G$
  2. Also we have $P_5\cong \mathbb Z_{25}$ since there is only one group of order $25$.
  3. 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.

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    $\begingroup$ There are even $4$ abelian groups...as for each $p=2,5$ you can take either the cyclic or non-cyclic group of order $p^2$. $\endgroup$ – lulu Jan 9 '18 at 16:45
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    $\begingroup$ I know. This doesn't answer my question though. $\endgroup$ – Buh Jan 9 '18 at 16:47
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    $\begingroup$ Ok, so you learned that given two groups $H$ and $N$ there is no unique semidirect product $N\rtimes H$. To give you an example there are four non-isomorphic semidirect products of $C_{15}$ and $C_2$. There's the direct product $C_{15}\times C_2$. A semidirect product where conjugation by the generator of $C_2$ is trivial on $C_3$ ut non-trivial on $C_5$. This group is really $C_3\times(C_5\rtimes C_2)\simeq C_3\times D_5$. Then there is the variant where the above conjugation is trivial on $C_5$ and non-trivial on $C_3$ yielding $(C_3\rtimes C_2)\times C_5\simeq S_3\times C_5$ $\endgroup$ – Jyrki Lahtonen Jan 9 '18 at 18:03
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    $\begingroup$ (cont'd) and lastly the case where the conjugation is non-trivial for both factors of $C_{15}=C_3\times C_5$, giving rise to $D_{15}$. More precisely, a semidirect product $N\rtimes H$ really should be written as $N\rtimes_\varphi H$, where $\varphi$ is a homomorphism from $H$ to $Aut(N)$ describing the conjugations of elements of $N$ by elements of $H$. Different choices of $\varphi$ may lead to non-isomorphic semidirect products. The direct product corresponds to the case where $\varphi$ is the trivial homomorphism $\varphi(h)=id_N$ for all $h\in H$. $\endgroup$ – Jyrki Lahtonen Jan 9 '18 at 18:05
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For starters: There's at least two groups of order 25; $\mathbb{Z}_{25}$ and $\mathbb{Z}_5 \times \mathbb{Z}_5$. To see that they're different, note that the first has an element of order 25, and the latter only has elements of order 5 or less.

Another way you're wrong is that you dismiss the non-trivial semi-direct products of $P_2$ and $P_5$. What automorphisms of $P_{25}$ are there? There's more than just the trivial one.

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  • $\begingroup$ I don't understand too much of the semi-direct product. In fact, I just applied the theorem: Let $H,N$ be subgroups of $G$ where $N$ is normal such that $N\cap H=1$ and $G=HN$. Then $G\cong H\ltimes N$. Somehow I can't apply this theorem because I am missing something or I misunderstand it. $\endgroup$ – Buh Jan 9 '18 at 17:09
  • $\begingroup$ @Buh You're missing something, namely, what the definition of the semi-direct product is. For a concrete example, consider $D_{8}$, the dihedral group on $4$ vertices. You can show that $D_{8} = \mathbb{Z}_4 \ltimes \mathbb{Z}_2$, but $D_{8} \neq Z_2 \times Z_4$, since it's not Abelian. Namely, what you're missing is that the semi-direct product gives a way for the factors to "see" and "interact" with each other. $\endgroup$ – Duncan Ramage Jan 9 '18 at 17:17
  • $\begingroup$ Hmm, not quite clear enough for me. So let's say we have $n$ possibilities for a group $N$ and $m$ possibilities for a group $H$ in this setup. Then $H\ltimes N$ doesn't have $nm$ possibilites? $\endgroup$ – Buh Jan 9 '18 at 17:21
  • $\begingroup$ It doesn't necessarily have $nm$ possibilities. It may still be the case that there's $nm$ possibilities, but there may be more. $\endgroup$ – Duncan Ramage Jan 9 '18 at 17:23
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    $\begingroup$ Wow, I really don't understand how that can be. But I'll just leave it at this for now. Thanks for pointing that out, wouldn't have thought about this in a million years. $\endgroup$ – Buh Jan 9 '18 at 17:29
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To start, there are two groups of order $25$, $\mathbb{Z}_{25}$ and $\mathbb{Z}_5 \times \mathbb{Z}_5$

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  • $\begingroup$ That's why I wrote "Obviously this is false" in my post. Since everybody got this wrong I will highlight it more. $\endgroup$ – Buh Jan 9 '18 at 16:48
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    $\begingroup$ @Buh I don't see how "everybody got this wrong", assuming that there's only one group of order 25 is the mistake in your proof. $\endgroup$ – Duncan Ramage Jan 9 '18 at 16:51
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    $\begingroup$ I misread your comment, sorry. So let me rephrase my question here: There are two groups of order $25$ (since groups of order $p^2$ must be abelian), this yields 4 groups of order $100$ in my case. In groupprops, however, it says that there are 16 groups of order $100$ up to isomorphy. $\endgroup$ – Buh Jan 9 '18 at 16:52
  • $\begingroup$ @Bah Good point. I expanded on this in my answer. $\endgroup$ – Duncan Ramage Jan 9 '18 at 17:06

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