# Thread: TA’s Fun Algebra Problem

1. ## TA’s Fun Algebra Problem

Hi all.

Following NonCommAlg’s example, I have decided to post a fun algebra (group theory) problem of my own on this forum.

Let $K=G\times H,$ where $G$ is an abelian group and $H$ is a cyclic group of order 2, and suppose $G$ contains an element $g$ with $g^2\ne1_G.$ Find a binary operation $\circ$ on $K$ such that $\left$ is a nonabelian group.

To NonCommAlg: If you know the answer to this one (and I have a feeling you do) please don’t be too quick to reveal the answer. Let others have a chance to do the problem first.

2. Originally Posted by TheAbstractionist
Hi all.

Following NonCommAlg’s example, I have decided to post a fun algebra (group theory) problem of my own on this forum.
Let $K=G\times H,$ where $G$ is an abelian group and $H$ is a cyclic group of order 2, and suppose $G$ contains an element $g$ with $g^2\ne1_G.$ Find a binary operation $\circ$ on $K$ such that $\left$ is a nonabelian group.
To NonCommAlg: If you know the answer to this one (and I have a feeling you do) please don’t be too quick to reveal the answer. Let others have a chance to do the problem first.
it took me some time to figure this one out!

Spoiler:
let $H=\{1, x \}.$ define $\theta: H \longrightarrow \text{Aut}(G)$ by $\theta(1)=\text{id}_G$ and $\theta(x)(a)=a^{-1}, \ \forall a \in G.$ now define $\circ$ by $(g_1,h_1) \circ (g_2,h_2)=(g_1 \theta(h_1)(g_2), h_1h_2), \ \ \forall g_1,g_2 \in G$ and $\forall h_1,h_2 \in H.$

then $$ is a group because $ \cong G \rtimes_{\theta} H$ and it's non-abelian because for any $g \in G$ with $g^2 \neq 1$ we have: $(1,x) \circ (g,1) \neq (g,1) \circ (1,x).$

3. Originally Posted by NonCommAlg
it took me some time to figure this one out!

Spoiler:
let $H=\{1, x \}.$ define $\theta: H \longrightarrow \text{Aut}(G)$ by $\theta(1)=\text{id}_G$ and $\theta(x)(a)=a^{-1}, \ \forall a \in G.$ now define $\circ$ by $(g_1,h_1) \circ (g_2,h_2)=(g_1 \theta(h_1)(g_2), h_1h_2), \ \ \forall g_1,g_2 \in G$ and $\forall h_1,h_2 \in H.$

then $$ is a group because $ \cong G \rtimes_{\theta} H$ and it's non-abelian because for any $g \in G$ with $g^2 \neq 1$ we have: $(1,x) \circ (g,1) \neq (g,1) \circ (1,x).$
Hi NonCommAlg.

Yes, that was what I had in mind.

For those who don’t understand NonCommAlg’s proof:
Spoiler:
The binary operation is this: For $g_1,g_2\in G,\ H=\left,$

$\left(g_1,1_H\right)\circ\left(g_2,1_H\right)\ =\ \left(g_1g_2,1_H\right)$

$\left(g_1,1_H\right)\circ\left(g_2,h\right)\ =\ \left(g_1g_2,h\right)$

$\left(g_1,h\right)\circ\left(g_2,1_H\right)\ =\ \left(g_1g_2^{-1},h\right)$

$\left(g_1,h\right)\circ\left(g_2,h\right)\ =\ \left(g_1g_2^{-1},1_H\right)$

Spoiler:
You can verify that all the group axioms are satisfied for the operation defined above, but checking associativity directly can be rather time consuming. Fortunately, we have a theorem.

Let $G$ be any group, $H=\left$ a cyclic group of order $r$ and $\vartheta$ an automorphism of $G$ such that $\vartheta^r$ is the identity automorphism. Then $K=G\times H$ is a group with respect to the operation $\circ$ where

$\left(g_1,h^i\right)\circ\left(g_2,h^j\right)\ =\ \left(g_1\vartheta^i(g_2),h^{i+j}\right)$

for all $g_1,g_2\in G$ and $i,j\in\mathbb Z.$

$K$ is known as the cyclic extension of $G$ by $H$ induced by the automorphism $\vartheta.$ For a proof of the theorem (and more on cyclic extensions) see Chapter 21 of John F. Humphreys, A Course in Group Theory, Oxford University Press, 1996.

In our example, $\vartheta$ is the mapping $g\mapsto g^{-1}$ which is an automorphism because $G$ is Abelian. And $K$ is not Abelian because $(g,1)(g,h)=(g^2,h)$ whereas $(g,h)(g,1)=(1,h);$ the two are not equal if $g^2\ne1.$