I just had an exam, and I'm curious to see if I got it right, because my professor didn't do it like I did, and I didn't have time to hear his final answer.(adsbygoogle = window.adsbygoogle || []).push({});

Well, you tell me if I have any mistakes:

Let G be such a group.

147=3*7^2. Let P be the 7-Sylow subgroup (it is unique because it is of index 3, being the lowest prime that divides |G|, and is therefor normal), and let Q be a 3-Sylow.

It is obvious that their intersection is trivial, whereas G=PQ. Therefor, G is the half direct product of P and Q.

P being the normal subgroup, we want to find all non-trivial homomorphisms from Z_3 to Aut(Z_7xZ_7) or Aut(Z_49). The second one is simple, so let's look at

[tex]$Z_3 \to Aut\left( {Z_7 \times Z_7 } \right)$

[/tex]

I played with it and got the conclusion that by presenting

[tex]$Z_7 \times Z_7 = \left\langle a \right\rangle \times \left\langle b \right\rangle $

[/tex]

The only options where, if [tex]$Z_3 = \left\langle y \right\rangle $

[/tex]:

[tex]\[

\left\{ \begin{array}{l}

yay^{ - 1} = a, yay^{ - 1} = b^2 \\

yay^{ - 1} = a^2 , yay^{ - 1} = b \\

yay^{ - 1} = a^2 , yay^{ - 1} = b^2 \\

\end{array} \right.

\]

[/tex]

And the trivial direct product, of course.

What do you think? Did I get it right?

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# Classify all groups of order 147

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