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No. Consider the semi-direct product $(\mathbf{Z}\times \mathbf{Z}/2)\rtimes \mathbf{Z}/2$, where the rightmost factor acts by $(1,0)\mapsto (1,1)$ and fixes $(0,1)$. Then the normal closure of $\math …

Is there any sort of classification of (say finite) groups with the property that every subgroup is normal? Of course, any abelian group has this property, but the quaternions show commutativity isn' …

Let $G=\mathbb{R}^2$ with order defined by $(a_1,a_2)\leq (b_1,b_2) \iff a_1\leq b_1$ and $a_2\leq b_2$. Let $u=(1,1)$ and let $g_0=(1,2)$. Then $p(g_0)=1$ and $r(g_0)=2$.

By a well known lemma in group theory, we can write any $g\in G$ uniquely as $h_1h_2$ where $h_1$ has order a power of $p$, $h_2$ has order prime to $p$, and $h_1h_2=h_2h_1$. Thus we may write $G$ as …