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I am trying to determine what group this is a presentation of: $$\langle b,c \mid 2(b+c)=0, b+c=c+b \rangle.$$ I am pretty sure it is $\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$, but am stuck on how to show it.

Shaun
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Squeeze my theorem
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    $bc=cb$ means that $G$ is abelian. $G$ is one-relator so it's infinite. Then it's generated by $b$ of infinite order and $bc$, of order $2$. Furthermore, $\langle b\rangle$ and $\langle c\rangle$ intersect trivially. But you might want to make your presentation consistent: it's a mixture of additive and multiplicative notation. – David A. Craven Aug 05 '20 at 19:50
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    @DavidA.Craven Ah yes, I see now that 2(b+c)=0 just means that the element b+c has order 2. Then I could replace the generator c by b+c to get <b,b+c | 2(b+c)=0, b+c=c+b> and easily see that this is Z x Z/2Z. Thanks! – Squeeze my theorem Aug 05 '20 at 19:55

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Put multiplicatively, your presentation is

$$\langle b,c\mid (bc)^2, bc=cb\rangle.$$

By Tietze transformations, introduce $a=bc$, so that $c=b^{-1}a$, giving the presentation

$$\langle a,b\mid a^2, a=(b^{-1}a)b\rangle,$$

which simplifies to

$$\langle a,b\mid a^2, ba=ab\rangle;$$

and this presentation yields, multiplicatively, the group $\Bbb Z_2\times \Bbb Z$. This is isomorphic to $\Bbb Z\oplus \Bbb Z/2\Bbb Z$.

Shaun
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