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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.

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    $\begingroup$ It is true, and a good exercise too. $\endgroup$ – Olivier Bégassat Nov 4 '14 at 11:35
  • $\begingroup$ Are you sure @Olivier Bégassat $\endgroup$ – Shanaya Sharma Nov 4 '14 at 11:41
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    $\begingroup$ 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. $\endgroup$ – Gerry Myerson Nov 4 '14 at 12:33
  • $\begingroup$ Hint: as a 2-group G is nilpotent and so has a non trivial center. Consider how this center can be extended. $\endgroup$ – Marc Bogaerts Nov 4 '14 at 14:05
  • $\begingroup$ @Nimda, I have a feeling that someone asking a question at this level isn't going to know about nilpotent, center, or extending. $\endgroup$ – Gerry Myerson Nov 4 '14 at 22:11
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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 $$\begin{cases}|G| = 8\\G=\langle a,b\rangle\\a^4 = e\\b^2 = a^2\\ba = a^3b\end{cases}$$

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

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