Let $G$ be a group of order $8$ and $x$ be an element of $G$ of order $4$. Prove that $x^2 \in Z(G)$

Asked Apr 18 '18 at 16:49

Active Apr 18 '18 at 16:56

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I am quite a beginner in group theory, so I need to get my proofs checked. Here's what I've done:

We know, that the center of a group $G$, i.e $Z(G)$ is a normal subgroup of $G$ [no proof required here]. So, the quotient group $G/Z$ can be considered. Now, consider $x^2Z*x^2Z=x^4Z=Z=(x^2Z)^2$, which readily implies that $x^2 \in Z(G)$.

Is this all? Or am I missing something?

edited Apr 18 '18 at 16:56

asked Apr 18 '18 at 16:49

Subhasis Biswas

2,600

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You did not use that $ord(x)=4$? – Dietrich Burde Apr 18 '18 at 16:53
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How does $xZ\cdot xZ=x^2Z$ imply that $x^2\in Z$? This holds for all $x$ in all groups. Yet, there are many groups where $x^2$ is not necessarily an element of the center. At some point you must use the facts that $x$ has order four, and $G$ has order eight. – Jyrki Lahtonen Apr 18 '18 at 16:54
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I have no idea what you mean. How does $xZ\cdot xZ=x^2Z$ imply $x^2\in Z$? – tomasz Apr 18 '18 at 16:54
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Hint: If $G/Z$ is cyclic then $G$ is abelian (prove that if you haven't seen it before). – Jyrki Lahtonen Apr 18 '18 at 16:56
I have edited the body, is it correct now? – Subhasis Biswas Apr 18 '18 at 16:57
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No. Still not correct. You aren't even trying to justify the claim $(x^2Z)^2=Z\implies x^2\in Z$. The claim is false in general. You need to use the fact that $G$ has order eight in an essential way. – Jyrki Lahtonen Apr 18 '18 at 16:58
can I proceed with the claim though? Is there any hope? – Subhasis Biswas Apr 18 '18 at 17:00
You have to prove that $xZ$ has order two in $G/Z$. One route is to prove that A) $Z(G)$ cannot be trivial, B) $G/Z(G)$ cannot be cyclic. There may be others. – Jyrki Lahtonen Apr 18 '18 at 17:03
Since the factors of $8$ are $1, 2, 4, 8$ and $o(Z(G))=8$ readily solves the problem. So, the rest of the cases are $1, 2, 4$. – Subhasis Biswas Apr 18 '18 at 17:04
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A totally elementary solution is contained in this old answer of mine. For the other route, see for example this. – Jyrki Lahtonen Apr 18 '18 at 17:10
Let us continue this discussion in chat. – Subhasis Biswas Apr 18 '18 at 17:25

Let $G$ be a group of order $8$ and $x$ be an element of $G$ of order $4$. Prove that $x^2 \in Z(G)$

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