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Let $G$ be a group whose presentation is $G=\{x,y| x^5=y^2=e, x^2y=yx\}$ Then $G$ is isomorphic to:

$(a)\mathbb Z_5$ (b) $\mathbb Z_{10}$ (c) $\mathbb Z_2$ (d) $\mathbb Z_{30}$.

The correct option is (c). But I get more than 2 elements in G because at least powers of x like x,x^2, x^3,x^4 must be in G, then how it is isomorphic to z2. I am very poor in Abstract Algebra, somebody help me please! I will be so much thankful to you if you explain in a little bit more detail.

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    If $G$ is isomoriphic to $\mathbb Z_2,$ it must be the case that $x = e.$ So, assume that $x\ne e$ and create a contradiction. – 2017-01-25
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    How is that a contradiction? The group $\mathbb{Z}_2$ also contains the powers $g^2, g^3, g^4, \dots$ of its generator $g$. – 2017-01-25
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    if you prove $x=e$ then you only get the elements $y$ and $y^2$ because the order of $y$ is at most two. – 2017-01-25

2 Answers 2

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To address your concern about $x,x^2,x^3,x^4$, note that there is no restriction that they are four distinct group elements. For instance, $x$ could be the identity $e$ and still satisfy $x^5=e$, and the group contains $x^k=e$ for all $k$.

Using $y^2=e$ and $x^2 y = yx$ we have $$x = y^{-1} x^2 y = yx^2 y = y(yx^2 y)(yx^2 y) y = x^4.$$ Combining this with $x^5=e$ will imply $x=e$.

So, $G=\{y \mid y^2 = e\} = \mathbb{Z}_2$.

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We know $$x^5=e$$ $$y^2=e$$ $$x^2y=yx$$

This implies $$xy=y^2xy=yyxy=yx^2yy=yx^2y^2=yx^2$$

$$x^4y=x^2x^2y=x^2yx=yxx=yx^2$$

Hence, we have $x^3=e$ and because of $x^5=x^3$ we can also conclude $x^2=e$. Finally, we have $x=e$. This shows that $G$ has only $2$ elements.