Classify all groups of order 147

[Palindrom]

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.

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

I played with it and got the conclusion that by presenting
$$Z_7 \times Z_7 = \left\langle a \right\rangle \times \left\langle b \right\rangle$$
The only options where, if $$Z_3 = \left\langle y \right\rangle$$:
$$\[ \left\{ \begin{array}{l} yay^{ - 1} = a, yay^{ - 1} = b^2 \\ yay^{ - 1} = a^2 , yay^{ - 1} = b \\ yay^{ - 1} = a^2 , yay^{ - 1} = b^2 \\ \end{array} \right. \]$$
And the trivial direct product, of course.
What do you think? Did I get it right?

 

[AKG]

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.

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

I played with it and got the conclusion that by presenting
$$Z_7 \times Z_7 = \left\langle a \right\rangle \times \left\langle b \right\rangle$$
The only options where, if $$Z_3 = \left\langle y \right\rangle$$:
$$\[ \left\{ \begin{array}{l} yay^{ - 1} = a, yay^{ - 1} = b^2 \\ yay^{ - 1} = a^2 , yay^{ - 1} = b \\ yay^{ - 1} = a^2 , yay^{ - 1} = b^2 \\ \end{array} \right. \]$$
And the trivial direct product, of course.
What do you think? Did I get it right?

Uh, should that second column have b-conjugates? I don't know how you're getting what you're getting. However, I'm quite sure that there is more than just the trivial automorphism.

The group Aut(Z7 x Z7) is isomorphic to GL(2,Z7). I think, in general, Aut(Fⁿ) is isomoprhic to GL(n,F). So it suffices to find the 2 x 2 matrices over the field Z7 with non-zero determinant. This is equivalent to finding a basis of the vector space Z7 x Z7. There are 49-1 choices for the first vector in your basis (any vector but 0). For your second choice, you must choose something linearly independent from the first. For the second, you can choose anything that's not in the span of the first, so there's 49 - 7. Total: 48*42 = 2⁵3²7.

So Aut(Z7 x Z7) does indeed have a subgroup of order 3, hence is at least one non-trivial semi-direct product between Z3 and Z7 x Z7.

 

[Palindrom]

Of course there is, that's what my last paragraph is all about. I found three non trivial such products.
I didn't say that there was only the trivial automorphism. I don't understand what you meant, if you could please explain... Do you not agree that the three groups that I wrote are good?

And you're right, the second column should have b's instead of a's in the middle.

 

[AKG]

You haven't shown that your three options will lead to non-isomorphic groups, and that there are no other options.

 

[Palindrom]

Oh no I have, I just didn't write everything here.

By the way, there is one more option with Z_49, but as I said that's really simple so I didn't get into it.

I'm just asking if the three ones that I found are correct.