# Does this uniquely determine G?

• Oct 9th 2011, 03:41 AM
Deveno
Does this uniquely determine G?
(up to isomorphism)

$\displaystyle 1\longrightarrow \mathbb{Z}_4 \longrightarrow^{\!\!\!\!\!\!\!\!\!\beta}\ \, G \longrightarrow^{\!\!\!\!\!\!\!\!\!\alpha}\ \, \mathbb{Z}_2 \longrightarrow 1$

$\displaystyle \mathbb{Z}_2\longrightarrow^{\!\!\!\!\!\!\!\gamma} \ \, G$

$\displaystyle \gamma \circ \alpha = 1_{\mathbb{Z}_2}$

(if my latex were better i'd try to put it all on one line, this is a right-split short exact sequence).

if so, is there a "smaller" diagram which conveys the same information?
• Oct 9th 2011, 10:08 AM
Drexel28
Re: Does this uniquely determine G?
Quote:

Originally Posted by Deveno
(up to isomorphism)

$\displaystyle 1\longrightarrow \mathbb{Z}_4 \longrightarrow^{\!\!\!\!\!\!\!\!\!\beta}\ \, G \longrightarrow^{\!\!\!\!\!\!\!\!\!\alpha}\ \, \mathbb{Z}_2 \longrightarrow 1$

$\displaystyle \mathbb{Z}_2\longrightarrow^{\!\!\!\!\!\!\!\gamma} \ \, G$

$\displaystyle \gamma \circ \alpha = 1_{\mathbb{Z}_2}$

(if my latex were better i'd try to put it all on one line, this is a right-split short exact sequence).

if so, is there a "smaller" diagram which conveys the same information?

Presumably you meant $\displaystyle \alpha\circ\gamma=\text{id}_{\mathbb{Z}_2}$. If so, we can definitely say that $\displaystyle G\cong \mathbb{Z}_4\rtimes_\varphi\mathbb{Z}_2$ for some homomorphism $\displaystyle \varphi:\mathbb{Z}_2\to\text{Aut}(\mathbb{Z}_4)$. But, since $\displaystyle \text{Aut}(\mathbb{Z}_4)=\mathbb{Z}_2$ there is precisely one non-trivial homomorphism. And so, in particular, up to isomorphism the only semidirect product gives $\displaystyle D_4$. Thus, $\displaystyle G\cong D_4$.
• Oct 9th 2011, 11:35 AM
NonCommAlg
Re: Does this uniquely determine G?
$\displaystyle G \cong \mathbb{Z}_4 \times \mathbb{Z}_2$ satisfies the conditions too. so $\displaystyle G$ is not uniquely determined.
• Oct 9th 2011, 11:36 AM
Drexel28
Re: Does this uniquely determine G?
Quote:

Originally Posted by NonCommAlg
$\displaystyle G \cong \mathbb{Z}_4 \times \mathbb{Z}_2$ satisfies the conditions too. so $\displaystyle G$ is not uniquely determined.

Right, I forgot the trivial homomorphism $\displaystyle \varphi:\mathbb{Z}_2\to\mathbb{Z}_2$.
• Oct 9th 2011, 12:19 PM
Deveno
Re: Does this uniquely determine G?
yes i did mean the other composition ( can i claim it was because i was looking at herstein last night? no? ok, i admit it, i'm just stupid).

so...how would i indicate there is no left-split with a picture, to rule out the trivial semi-direct (that is direct) product?
• Oct 9th 2011, 12:23 PM
Drexel28
Re: Does this uniquely determine G?
Quote:

Originally Posted by Deveno
yes i did mean the other composition ( can i claim it was because i was looking at herstein last night? no? ok, i admit it, i'm just stupid).

so...how would i indicate there is no left-split with a picture, to rule out the trivial semi-direct (that is direct) product?

You mean you just want to indicate, diagramatically that there is no left split?
• Oct 9th 2011, 12:28 PM
Deveno
Re: Does this uniquely determine G?
correct. what got me thinking about this, is thinking about identifying D4 solely in terms of homomorphisms to and from it (although i admit including the extra information of the short exact sequence is cheating a little) and wondering how well this can be done, for an arbitrary group.