Let $H$ be an abelian subgroup of index $2$ in $G$, and $G=\langle H,g \rangle$ then why is it only units of $\Bbb{Z}(G/H)$ are trivial (i.e. of the form $\pm\bar{g}$)?

An arbitrary element of $\Bbb{Z}(G/H)$ will be of the form $a_1+a_2 \bar{g}$ where $a_1,a_2 \in \Bbb{Z}$. But if it is a unit why does it have to be of the form $\pm\bar{g}$?

  • $\begingroup$ What does Z(G/H) mean? the group ring? $\endgroup$ – quid Dec 28 '15 at 16:46
  • $\begingroup$ yes... integral group ring of factor group G/H $\endgroup$ – Bhaskar Vashishth Dec 28 '15 at 16:46
  • $\begingroup$ Thanks. Had I seen the tag I might not have asked. $\endgroup$ – quid Dec 28 '15 at 16:48

Define a function $N:\mathbb{Z}(G/H)\to \mathbb{Z}$ by $N(a+b\overline{g}):=a^2-b^2$. Notice that $N$ is multiplicative, that is to say, $$N((a+b\overline{g})(c+d\overline{g})) = N(a+b\overline{g})N(c+d\overline{g}).$$ From there, it follows that if $a+b\overline{g}$ is a unit, then $N(a+b\overline{g})=a^2-b^2$ in invertible in $\mathbb{Z}$. In other words, $a^2-b^2= \pm 1$. Since $a$ and $b$ are integers, this is only possible if $a=\pm 1$ and $b=0$, or $a=0$ and $b=\pm 1$. Hence the result.


Let me write $g$ instead of $\overline {g}$. One has

$(a + bg)(c + d g) = ac + bd + (ad + bc)g$.

So for $a+ b g$ to be a unit you need $c,d$ such that $ac + bd = 1$ and $ad + bc = 0$.

If $a=0$ you get what you want directly. Assume $a \neq 0$.

You get $a^2c + abd =a$ and as $ad= -bc$ you get $a^2c - b^2c =a$, whence $c(a+b)(a-b)= a$.

Considering the absolute value it follows that $|a+b|$ and $|a-b|$ are not greater than $|a|$ which is only possible if $b=0$. Further it follows $ca^2 = a$ so $ac=1$ (and $d=0$).

Thus the units are $\pm 1$ and $\pm g$. (Which seems slightly different from what you claim.)

This is a quite pedestrian solution but seems to work.

  • $\begingroup$ yes $\pm1$ are in obvious trivial units. Sorry for not mentioning. This works. $\endgroup$ – Bhaskar Vashishth Dec 28 '15 at 17:07

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