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

Now as $$x$$ is of order $$4$$ and it generates $$H=\langle x\rangle$$ a cyclic subgroup of $$G$$ of order $$4$$, hence index of $$H$$ in $$G$$ is $$2$$. So $$H=\langle x\rangle$$ is normal in $$G$$.

Now $$gH=Hg$$, for all $$g$$ in $$G$$.

=> $$g(x^2)=(x^2)g\:$$ (since $$x^2$$ is an element of $$H$$), for all $$g$$ in $$G$$.

Hence $$x^2 \in Z(G)$$.

This was my approach. (I think that I am wrong in here, $$g(x^2)=(x^2)g$$) Please, correct me wherever necessary. And if you could provide a better solution, please do.

Hint:

Normality only implies that $$gH=Hg$$, i.e. $$gh_1 = h_2g$$ where $$h_1,h_2\in H$$ might be different. Since $$x$$ has order four, list all elements of $$H$$. If $$gx^2 = hg$$, what could $$h$$ be?

– Rkb
May 8, 2019 at 12:56

$$gH=Hg$$ does not imply that $$gh=hg$$ for all $$h\in H$$. These are different things.

Here is a possible solution. Let $$g\in G$$. Then $$gx^2g^{-1}=(gxg^{-1})^2$$. Since $$\langle x\rangle\trianglelefteq G$$ we know that $$gxg^{-1}\in\langle x\rangle$$. Now what $$gxg^{-1}$$ might be? Of course it can't be the identity $$e$$ because the identity is conjugate only to itself. Also it can't be $$x^2$$ because then we would get $$e=(x^2)^2=(gxg^{-1})^2=gx^2g^{-1}$$ which is again a contradiction. Hence $$gxg^{-1}$$ is either $$x$$ or $$x^3$$ and hence $$gx^2g^{-1}=(gxg^{-1})^2=x^2$$.

This can be obtained by several methods.

(1) There are only two non-abelian groups of order $$8$$: $$D_8$$ and $$Q_8$$. You can check them respectively.

(2) $$Z(G)$$ must have order $$p$$ if $$G$$ has order $$p^3$$, with $$G/Z(G)\cong\mathbb{Z}_p\times\mathbb{Z}_p$$. In this case $$p = 2$$ and so $$G/Z(G)\cong\mathbb{Z}_2\times\mathbb{Z}_2$$. Hence for any $$gZ(G)\in G/Z(G)$$, $$g^2Z(G) = Z(G)$$.

Therefore, a more general case is that if $$G$$ has order $$p^3$$, where $$p$$ is a prime number, then $$g^p\in Z(G)$$ for all $$g\in G$$ (not necessarily to be of order $$p^2$$). In your question, $$x$$ is not necessarily to be of order $$4$$.

A priori, you don't have $$gx^2=x^2g$$, but what you have is $$gx^2=hg$$ for some $$h\in H$$. So, you have $$gx^2g^{-1}\in H=\{1,x,x^2,x^3\}$$. So, what you actually have is $$gx^2g^{-1}=x^i,\hbox{ for some }i=0,1,2,3$$ Now, the order of an element is preserved by conjugation,i.e. $$o(h)=o(ghg^{-1})$$(this is because conjugation is an isomorphism). In particular $$o(x^i)=o(gx^2g^{-1})=o(x^2)=2.$$But $$o(x^0)=1$$ and $$o(x^1)=o(x^3)=4$$. So, the only possibility is $$i=2$$. Hence $$gx^2g^{-1}=x^2$$

Your approach is not wrong at all , but it needs some clarity. And also a better understanding of what have u constructed. So,

As u have shown, $$\Bbb{H}$$= $$$$ , where o(x)=4. And as $$\Bbb{H}$$ , is cyclic , order of it is also 4 . And [ $$\Bbb{S}$$ : $$\Bbb{H}$$ ]=2, thus $$H\trianglelefteq S$$ . Now as identity is unique and it is in $$\Bbb{H}$$ , then no element from $$\Bbb{S}$$ \ $$\Bbb{H}$$ will have order 1 .

Now when we are considering g, it can be picked up either from $$\Bbb{H}$$ or from , $$\Bbb{S}$$ \ $$\Bbb{H}$$.

Now if it's taken from $$\Bbb{H}$$ , as $$\Bbb{H}$$ is cyclic, $$\Bbb{H}$$ must be commutative, thus ,

$$\forall$$ g in $$\Bbb{H}$$ , g $$x^{2}$$ = $$x^{2}$$g holds.

Now when g is in $$\Bbb{S}$$ \ $$\Bbb{H}$$, o(g)≠1 . If it is 4 then the same argument comes . Now if o(g)=2=o( $$x^{2}$$ ), then they will commute and ,

g $$x^{2}$$= $$x^{2}$$ g holds .

And at last if, o(g)=8 , Then as o(g)=o( $$\Bbb{S}$$ ) , $$\Bbb{S}$$ will be cyclic. Thus it will be commutative. And then also,

g $$x^{2}$$ = $$x^{2}$$g will hold.

So , $$\forall$$ g in $$\Bbb{S}$$ , the relation will hold .

Well, to understand this ongoing argument more clearly, let's have an example. Let's consider the dihedral group $$D_{4}$$ which has 8 elements and they are , {r, $$r^{2}$$ , $$r^{3}$$ , $$r^{3}$$ , b, b', d, d '} . Here,

b and b ' represent reflection about diagonals . d and d ' represent reflection about the lines joining middle points of opposite arms of the square .

r= $$R_{90}$$ , $$r^{2}$$= $$R_{180}$$ , $$r^{3}$$= $$R_{270}$$ , $$r^{4}$$= $$R_{360}$$ .

Now as, b=r $$\ast$$ b $$\ast$$ $$r^{-1}$$ And, d= r $$\ast$$ d' $$\ast$$ $$r^{-1}$$ . We can construct 2 subgroups of $$D_{4}$$ , which are ,

$$\Bbb{T}$$ = { e, b, b', $$r^{2}$$ }

$$\Bbb{F}$$ = { e, d, d' , $$r^{2}$$ }

Now , $$\Bbb{T}$$ $$\trianglelefteq$$ $$D_{4}$$

And , $$\Bbb{F}$$ $$\trianglelefteq$$ $$D_{4}$$

And most astonishingly these two are cyclic groups also . Additive cyclic groups , having the generator , $$r^{2}$$.

Now there leaves only two proper subgroups of $$D_{4}$$ , one is the group contained first four elements i.e {r, $$r^{2}$$ , $$r^{3}$$ , $$r^{3}$$ } , Which we have discussed earlier. So there left the group only {e, $$r^{2}$$ } .

And from all these one thing that we are sure about is , $$r^{2}$$ is only the element ( of order 2) which commutes with every element of $$D_{4}$$

So, $$r^{2}$$ $$\in$$ $$\Bbb{Z}$$( $$D_{4}$$ )

And it ends here .......( Well, it's just the beginning ! )....

The group $$G$$ looks like this: $$\{1,x,x^2,x^3,g,gx,gx^2,gx^3\}$$ Furthermore, $$g^2$$ must belong to $$\langle x\rangle$$. If $$g^2=x$$ or $$x^3$$ then $$G$$ is cyclic. If $$g^2=x^2$$ then $$x^2(gx^k)=g^3x^k=gx^2x^k=(gx^k)x^2$$

With a bit more sophistication and almost hands-down: let $$H=\langle x \rangle$$, where $$x$$ has order $$4$$. Then, $$|G:H|=2$$, so $$H$$ is normal and $$H \subseteq C_G(x) \subseteq G$$. It follows that $$G=C_G(x)$$ or $$H=C_G(x)$$. The first case is equivalent to $$x \in Z(G)$$ and then certainly $$x^2 \in Z(G)$$.
So let's assume $$H=C_G(x)$$. Then the $$G$$-conjugacy class $$|Cl_G(x)|=|G:C_G(x)|=2$$. Since in general a normal subgroup is the disjoint union of $$G$$-conjugacy classes, and $$1 \in N$$, this leads to $$x^3 \in Cl_G(x)$$ (remember that all the elements of a conjugacy class have the same order!) and thus $$Cl_G(x^2)=\{x^2\}$$. This is equivalent to $$x^2 \in Z(G)$$.

In general, if $$N \unlhd G$$ and $$|N|=4$$ then $$x^2 \in Z(G)$$ for all $$x \in N$$.

• Could you please explain, how $G=C_G(x)$ or $H=C_G(x)$ ?
– Rkb
May 10, 2019 at 12:45
• In general, if $H$ is a subgroup of $K$, both being subgroups of $G$ , then $|G:H|=|G:K| \cdot |K:H|$. So if the index $|G:H|$ is prime, you only have to choices: $H=K$ or $K=G$. May 10, 2019 at 13:30