# Given three non abelian group of order 8,two must be isomorphic.

True or false:

Given three non abelian group of order $8$,two must be isomorphic.

Solution:

Theorem : A non-Abelian group of order $8$ is isomorphic either to $D_4$ or $Q_8$.

I Think it is true.

• It is true, and a good exercise too. – Olivier Bégassat Nov 4 '14 at 11:35
• Are you sure @Olivier Bégassat – Shanaya Sharma Nov 4 '14 at 11:41
• Well, I concur with @Olivier; it is true, and it is a good exercise. Don't let anyone spoil your fun by doing it for you. – Gerry Myerson Nov 4 '14 at 12:33
• Hint: as a 2-group G is nilpotent and so has a non trivial center. Consider how this center can be extended. – Marc Bogaerts Nov 4 '14 at 14:05
• @Nimda, I have a feeling that someone asking a question at this level isn't going to know about nilpotent, center, or extending. – Gerry Myerson Nov 4 '14 at 22:11

Hint:

Show that if $G$ has an element of order $4$ and doesn't have an element of order 8, then $$\begin{cases}|G| = 8\\G=\langle a,b\rangle\\a^4 = e\\b^2 = a^u\\ba = a^sb\end{cases}$$

Where $u = 0,2$ and $s = 1,3$.

The first part comes by considering $H = \langle a\rangle$ and take $b \in G- H$ and consider the group $K$ spanned by $a$ and $b$, (Use Lagrange's Theorem.)

To define $u,s$ check the possibilities for $u,s \in \lbrace0,1,2,3\rbrace$ and remember that G does not have an element of order $8$.

The cases you want are $$\begin{cases}|G| = 8\\G=\langle a,b\rangle\\a^4 = e\\b^2 = e\\ba = a^3b\end{cases}$$ and $$\begin{cases}|G| = 8\\G=\langle a,b\rangle\\a^4 = e\\b^2 = a^2\\ba = a^3b\end{cases}$$

• @ShanayaSharma So did you get to prove that? – Aaron Maroja Nov 7 '14 at 21:02