Isomorphism in certain groups.

Question

Let $G = \langle \Bbb Z^3_2,+\rangle $, where as $ \Bbb Z_2= \{0,1\} $, $ \Bbb Z^3_2 = \{(x,y,z)\mid x,y,z \in \{0,1\}\} $, and the operation in $ \Bbb Z^3_2 $ is defined by $ (x_1,y_1,z_1) + (x_2,y_2,z_2) = (x_1+x_2,y_1+y_2,z_1+z_2) $ and the adding in $\Bbb Z_2 $ is defined by $\bmod 2$.

Let there be $H$ a group in order $8$ and $e$ the identity element, so that every $ x^2 = e, x \in H $. Prove that $H$ Isomorphic to $G=\langle \Bbb Z^3_2,+\rangle $

I know that i need to find a function $f$ so that $ G \to H \qquad \text{to every}\quad a,b \in G \quad f(a*b) = f(a)\circ f(b) \qquad $ and also $f$ have to be Surjective and injective.

How do I find that $f$?

What's my way of thinking?

Thanks !!

Answer 1

You could show that both of these are a vector space over $\mathbb{Z}/2$ and must be isomorphic because they have the same dimension.

Answer 2

The group $H$ is abelian, since $(xy)^2=1$ and $x^2=y^2=e$ for all $x,y\in H$.

Now, let $a,b$ be two distinct elements of $H$, chose any different element $c\in H$ from $e,a,b,ab$. We claim that $H=\{e,a,b,c, ab, ac, bc, abc\}$. This follows from the fact these elements are all different, for example, if $ab=bc$. We then have $ab=cb$, which implies $a=c$, by cancellation law. In fact, $H$ has the following presentation by generators and relations; $$H=\langle a,b,c | \space a^2=b^2=c^2=e,ab=ba,bc=cb,ac=ca\rangle$$

Now define the map $$\phi :\Bbb Z^3_2\to H$$ by $\phi((1,0,0))=a,\phi((0,1,0))=b,\phi((0,0,1))=c$. One can check that this map preserves the generators and the relations of $\Bbb Z^3_2$. Therefore, it is the precise isomorphism between them.

Further work for you if you would like it !

Try to show that any group of order 8 has to be isomorphic to one of the following groups;

$$\Bbb Z_8, \Bbb Z_4\times \Bbb Z_2, \Bbb Z^3_2, \mathrm{ Q}, \mathrm { D}_4 $$ Where the last two groups are Quaternion and Dihedral group respectively, which they are both non abelian groups.

Answer 3

Here's a functionless way to do it, though it is a little overkill. If $H$ is abelian, then by the structure theorem for abelian groups $H$ is isomorphic to $\mathbb{Z}_2^3$, $\mathbb{Z}_2\times \mathbb{Z}_4$, or $\mathbb{Z}_8$. It's easy to find elements of the latter two groups which have order greater than two, so it remains to prove that $H$ is not non-abelian. Since $H$ is a $2$-group, it has a non-trivial center, which by Lagrange's theorem has order $2$, $4$, or $8$. If the center has order $8$, then $H$ is abelian, and we are done. A center of a $p$-group cannot have index $p$, so the center cannot have order $4$. In the remaining case that $|Z(H)|=2$, $H/Z(H)$ cannot be cyclic, so $H/Z(H)\cong \mathbb{Z}_2^2$ by the structure theorem for abelian groups. Taking elements from the inverse image of a basis of $H/Z(H)$ in $H$ gives us (by assumption) two elements of $H$ of order two, and hence a subgroup of $H$ isomorphic to $\mathbb{Z}_2^2$. Since these commute with each other and the center, $H$ must then be $\mathbb{Z}_2^3$.