# A subgroup which is generated by $W$-marginal subgroups is itself $W$-marginal

There is a conclusion on $$W$$-marginal subgroups of a group:

A subgroup which is generated by $$W$$-marginal subgroups is itself $$W$$-marginal.

Here, $$W$$ is a set of words. In a group $$G$$, a normal subgroup $$N$$ is $$W$$-marginal in $$G$$ if $$w(g_1, \cdots, g_{i-1}, g_ia, g_{i+1}, \cdots, g_r) = w(g_1, \cdots, g_{i-1}, g_i, g_{i+1}, \cdots, g_r)$$ for all $$g_i \in G$$, $$a \in N$$ and all words $$w(x_1,x_2, \cdots, x_r)$$ in $$W$$. This is equivalent to the requirement: $$g_i \equiv h_i$$ mod $$N$$ $$(1 \leq i \leq r)$$, always implies that $$w(g_1, \cdots, g_r)=w(h_1, \cdots, h_r)$$.

But if I

Let $$W= \{ x^2 \}$$, $$G=D_4$$ (or sometimes denoted $$D_8$$), the dihedral group of order $$8$$. $$G$$ can be written as $$\langle a,b\mid a^2=b^4=1,abab=1\rangle= \{1, b, b^2, b^3, a, ab,ab^2, ab^3\}.$$ Let $$N_1=\{ 1, a, b^2, ab^2 \}$$, $$N_2=\{ 1, ab, b^2, ab^3 \}$$. Then $$N_1 \trianglelefteq G$$, $$N_2 \trianglelefteq G$$. Moreover, since both $$N_1$$ and $$N_2$$ consist of elements of order $$1$$ or $$2$$, $$N_1$$ and $$N_2$$ are $$W$$-marginal subgroups of $$G$$. $$N_1$$ and $$N_2$$ generate $$G$$, but $$G$$ is not $$W$$-marginal, as $$b\in G$$ and $$b^2 \neq 1$$.

I am eager to know where is wrong, and why the conclusion is all right. Thank you very much!

(Page 62, Derek J.S. Robinson, A Course in the Theory of Groups, GTM 80)

• $N_2$ is not W-marginal: $a\equiv b\mod N_2$ and yet $1=a^2\neq b^2$. – user641 Aug 10 '11 at 16:11
• @Steve D: Thank you very much for pointing out. I considered only the case when one of the two elements equalled to $1$. – ShinyaSakai Aug 11 '11 at 13:01

Note that usually we define the (full) marginal subgroup by requiring that insertion of $a$, on either side, not affect the value of the word. This does not require the assumption that the subgroup be normal ahead of time. With the added assumption that your elements form a normal subgroup, this is equivalent, of course, since $Ng = gN$ as sets.

Theorem. Let $G$ be a group. The full marginal subgroup associated to $w=x^2$ is the set of central elements of exponent $2$; i.e., $w^*(G) = \{ g\in Z(G) \mid g^2 = 1\}$.

Proof. Let $g\in Z(G)$ be of exponent $2$. Then for all $x\in G$, $w(xg) = (xg)^2 = x^2g^2 = x^2$ (since $g$ is central and of exponent $2$), and likewise $w(gx) = (gx)^2 = g^2x^2 = x^2$. Thus, $g\in w^*(G)$.

Conversely, let $g\in w^*(G)$. Then $1 = w(1) = w(1g) = g^2$, so $g$ is of exponent $2$. Now let $x\in G$. Then we have that $w(gx) = w(x) = w(xg)$, hence $gxgx = x^2$. Thus, $gxg = x$, so $gx = xg^{-1} = xg$. Thus, $g$ centralizes $x$, so $g\in Z(G)$. QED

(More generally, the full marginal subgroup associated to $w(x) = x^n$ is the set of central elements of exponent $n$).

In particular, a sugroup of $G$ is marginal relative to $w=x^2$ if and only if it consists of central elements of order $2$. However, you are only checking that the elements be of order $2$.

(Added. It seems that you were only checking for elements such that $w(g)=1$; while this is necessary for $g$ to lie in the marginal subgroup, it is not sufficient.)

Thus, note that neither $N_1$ nor $N_2$ are $w$-marginal, since they both contain noncentral elements. The only $w$-marginal subgroups of the dihedral group of order $8$ are $\{1\}$ and $\{1,b^2\}$.

As to the proof that a subgroup generated by (normal) marginal subgroups is itself marginal: let $N_1$ and $N_2$ be normal $W$-marginal subgroups of $G$, where $W$ is a set of words. Then $N_1N_2$ is a normal subgroup, and is the subgroup generated by $N_1$ and $N_2$. Let $w\in W$ be a word in $k$ letters, let $n_1\in N_1$, $n_2\in N_2$, and let $g_1,\ldots,g_k\in G$. We need to show that $$w(g_1,\ldots,g_{i-1},g_i(n_1n_2),g_{i+1},\ldots,g_k) = w(g_1,\ldots,g_n).$$ Since $N_2$ is $W$-marginal, we have $$w(g_1,\ldots,g_{i-1},g_i(n_1n_2),g_{i+1},\ldots,g_k) = w(g_1,\ldots,g_{i-1},g_in_1,g_{i+1},\ldots,g_n).$$ Since $N_1$ is $W$-marginal, we have $$w(g_1,\ldots,g_{i-1},g_in_1,g_{i+1},\ldots,g_n) = w(g_1,\ldots,g_{i-1},g_i,g_{i+1},\ldots,g_n).$$ Thus, the required equality holds.

That is, for every $x\in N_1N_2$, for every $w\in W$, $$w(g_1,\ldots,g_{i-1},g_ix,g_{i+1},\ldots,g_k) = w(g_1,\ldots,g_n).$$ Inductively, the result holds for any finite family of marginal subgroups. If $\{N_i\}$ is an arbitrary family of marginal subgroups, and $x\in \langle N_i\rangle$, then there is a finite set $\{i_1,\ldots,i_k\}$ such that $x\in N_{i_1}\cdots N_{i_k}$, and thus $x$ is itself marginal by the finite case.

Thus, a subgroup generated by $W$-marginal subgroups is also $W$-marginal.

For some more on marginal and verbal subgroups, see this previous answer.

• Thank you very much. But I think there are still some points I don't understand. As you have pointed out, when the words are of the form $x^n$, the marginal subgroup is in the center. Does this hold when there are other types of words in $W$? Moreover, would you please give me some hints on the proof of the first statement "A subgroup which is generated by $W$-marginal subgroups is itself $W$-marginal"? Thanks a lot. – ShinyaSakai Aug 11 '11 at 13:18
• @ShinyaSakai: Not every marginal subgroup is necessarily central. The marginal subgroup associated to the word $[[x,y],z]$, for example, is the second center, which in a nilpotent group will properly contain the center. As for the proof of the statement, isn't it immediate? Use the verbal definition, not the congruence one, to show that if $g$ and $h$ are marginal elements, then so is $gh^{-1}$. – Arturo Magidin Aug 11 '11 at 16:11
• Splendid answer! Thanks you very much! – ShinyaSakai Aug 16 '11 at 2:58