Does this uniquely determine G?

**Deveno** · October 9th 2011, 03:41 AM

(up to isomorphism)

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

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

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

(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?

**Drexel28** · October 9th 2011, 10:08 AM

Originally Posted by Deveno

(up to isomorphism)

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

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

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

(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 $\alpha\circ\gamma=\text{id}_{\mathbb{Z}_2}$ . If so, we can definitely say that $G\cong \mathbb{Z}_4\rtimes_\varphi\mathbb{Z}_2$ for some homomorphism $\varphi:\mathbb{Z}_2\to\text{Aut}(\mathbb{Z}_4)$ . But, since $\text{Aut}(\mathbb{Z}_4)=\mathbb{Z}_2$ there is precisely one non-trivial homomorphism. And so, in particular, up to isomorphism the only semidirect product gives $D_4$ . Thus, $G\cong D_4$ .

**NonCommAlg** · October 9th 2011, 11:35 AM

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

**Drexel28** · October 9th 2011, 11:36 AM

Originally Posted by NonCommAlg

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

Right, I forgot the trivial homomorphism $\varphi:\mathbb{Z}_2\to\mathbb{Z}_2$ .

**Deveno** · October 9th 2011, 12:19 PM

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?

**Drexel28** · October 9th 2011, 12:23 PM

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?

**Deveno** · October 9th 2011, 12:28 PM

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.

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