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I was trying to classify groups of order 12 and I ended up with 5 different groups:

$\bullet$ $\Bbb{Z}_{12}$

$\bullet$ $\Bbb{Z}_2 \times \Bbb{Z}_6$

$\bullet$ $(\Bbb{Z}_2 \times \Bbb{Z}_2) \rtimes_{\alpha} \Bbb{Z}_3$ where $\alpha: (1,1) \rightarrow \bar{-1}$

$\bullet$ $(\Bbb{Z}_2 \times \Bbb{Z}_2) \times \Bbb{Z}_3$

$\bullet$ $\Bbb{Z}_3 \rtimes \Bbb{Z}_4$ where $\alpha$ sends the generator to $\bar{-1}$

I want to show that $(\Bbb{Z}_2 \times \Bbb{Z}_2) \times \Bbb{Z}_3 \cong D_{12}$ and $\Bbb{Z}_3 \rtimes \Bbb{Z}_4 \cong Q_{12}$.

My notes state that:

There is a homomorphism $f: D_{2n} \rightarrow G$ with f(a)=x and f(b)=y if and only if:

1) $x^2 = e$

2) $y^n = e$

3) $x^{-1}yx=y^{-1}$

If such a homomorphism exists, it is unique as $f(b^k) = y^k$ and $f(b^ka) = y^kx$. If |x|=2 and |y|=n f is 1-1.

There is a homomorphism $f:Q_{4n} \rightarrow G$ with f(a)=x and f(b)=y if and only if:

1) $x^4 = e$

2) $y^{2n} = e$

3) $x^{-1}yx = y^{-1}$

4) $x^2 = y^n$

If n>1, then f is 1-1 if and only if |x|=4 and |y|=2n.

Case 1: $(\Bbb{Z}_2 \times \Bbb{Z}_2) \times \Bbb{Z}_3 \cong D_{12}$

We know that:

1) $\bar{(1,1)} \in \Bbb{Z}_2 \times \Bbb{Z}_2$ has order 2

2) $\bar{1} \in \Bbb{Z}_3$ has order 3

3) $(\bar(1,1), \bar{-1})(\bar{(0,0)}, \bar{1}) = (\bar{(1,1)}+ \alpha(\bar{-1})(\bar{0,0}), 0 )$ But I'm not sure if I know what $\alpha(\bar{-1})(\bar{0,0})$ is. We know that $\alpha$ sends the generator to $\bar{-1}$ But I'm not sure how to actually do this calculation. What exactly is the operation, do we just add $\bar{-1}$ to $\bar{(0,0)}$ and then add $\bar{1}$ (since that's the inverse of $\bar{-1}$?)

However, we are also supposed to have $|\bar{(1,1)}|=n=6$, but that is not true in this case, right?

Case 2: $\Bbb{Z}_3 \rtimes \Bbb{Z}_4 \cong Q_{12}$

Let $\Bbb{Z}_3 = \langle b \rangle$ and $\Bbb{Z}_4 = \langle a \rangle$

1) $a^4 = e$

2) $b^6 = e$

3) $a^{-1}ba = b^{-1}$

4) Here I'm stuck, because we are supposed to have $a^2 = b^3 = e$ but that's not true, right?

I also have the same problem here about the order of b since $|b|=3 \not= 6$, so how can it be isomorphic to $Q_{4n}$?

Thank you in advance

user1729
  • 31,015

1 Answers1

5

Not quite right: your list of "five" groups of order $12$: There are indeed five groups of order $12$, but you've omitted one, and you've included a duplicate:

Note that $$(\mathbb Z_2 \times \mathbb Z_2) \times \mathbb Z_3 = \mathbb Z_2 \times (\mathbb Z_2 \times \mathbb Z_3) \cong \mathbb Z_2 \times \mathbb Z_6$$

since $\mathbb Z_2\times \mathbb Z_3 \cong \mathbb Z_6$, because $\gcd(2, 3) = 1$.

amWhy
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