While studying abelian groups, I came across the abelian group $$G = \frac{1}{4} \mathbb{Z} / \mathbb{Z}$$ what is this group? I've never seen this notation before?

  • $\begingroup$ Without knowing about the context I cannot make sure, but $\frac{1}{4}\mathbb{Z}$ is a subgroup of $\mathbb{Q}$ which contains $\mathbb{Z}$, so you can define the quotient group $\left(\frac{1}{4}\mathbb{Z}\right)/\mathbb{Z}$. There shouldn't be anything weird about the notation. $\endgroup$ – David Molano Apr 24 at 19:16

As @DavidMolano commented you can view $\frac{1}{4}\mathbb{Z}$ as a subgroup of $\mathbb{Q}$ containing $\mathbb{Z}$. Two elements $\frac{a}{4},\frac{b}{4}\in\frac{1}{4}\mathbb{Z}$ with $a,b\in\mathbb{Z}$ are congruent $\mod\mathbb{Z}$ iff $4|a-b$, and thus there is an isomorphism $\frac{1}{4}\mathbb{Z}/\mathbb{Z}\to\mathbb{Z}/4\mathbb{Z}$ defined as $\frac{a}{4}\to{}a\mod{}4$. In fact this is immediate, since every $a\in\frac{1}{4}\mathbb{Z}/\mathbb{Z}$ is congruent to $0,\frac{1}{4}, \frac{2}{4}$ or $\frac{3}{4}$.

  • $\begingroup$ "... are congruent mod Z ..." should be "are congruent mod 4" ? $\endgroup$ – mrblewog Apr 24 at 19:48
  • $\begingroup$ No, I think $\dfrac a 4 \equiv \dfrac b 4$ in $\frac14\mathbb Z/\mathbb Z$ iff $4|a-b$ (i.e., $a\equiv b \pmod 4$) $\endgroup$ – J. W. Tanner Apr 24 at 19:50

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