Classifying groups of order 8 without using conjugation

Question

I know that solutions to this exercise can be found quite easily but I have tried to solve it on my own and I'm stuck on the last case. I would be glad if someone could tell me if this is correct and how to continue:

The only case remaining is that $G$ has an element of order 4, say $g$. Then $\langle g\rangle$ is a normal subgroup.

Let's assume that there is an element $h$ of order 2 which is not $g^2$. Then $\langle h\rangle\cap\langle g\rangle=\{e\}$ and $\langle h\rangle\langle g\rangle=G$, so $G=\langle g\rangle\rtimes\langle h\rangle\cong\mathbb Z_4\rtimes\mathbb Z_2\cong D_4$.
Let's assume there is no such element $h$. So the only element of order 2 in $G$ is $g^2$. But then $G$ has 6 elements of order 4. Can I conclude that $G\cong Q_8$ from that when knowing that $Q_8$ has 6 elements of order 4 and one each of order 1 and 2? I'm not sure if having elements with the same order yields an isomorphism.

Edit: Maybe my question can be reduced to: Are there groups of the same order with all elements having identical order that are not isomorphic?

Yes there are, but not among the groups of order 8. See: If I know the order of every element in a group, do I know the group? — Ivan Neretin, Dec 01 '17 at 14:18
Ah, I see. Interesting. So is there a way to continue my proof without taking the "usual" way of cosets? — Buh, Dec 01 '17 at 14:26
Well, 8 is not all that much, so you might be able to do the trick simply by careful consideration of the Cayley tables. — Ivan Neretin, Dec 01 '17 at 14:30
Because I would like to see if one can obtain this result with a different approach. — Buh, Dec 01 '17 at 14:53

Dune · Accepted Answer · 2017-12-01T15:14:01.903

1

In order to construct an isomorphism $Q_8 \to G$, you first need to decide which definition of $Q_8$ you want to use. If you say e.g. $$ Q_8 := \langle i,j \mid i^4 = j^4 = 1, i^2=j^2, ij = ji^{-1} \rangle, $$ then you only need to find two generators $g,h \in G$ of order four satisfying $g^2 = h^2$ and $gh = hg^{-1}$. Another way of defining $Q_8$ (among many others) is $$ Q_8 := (C_4 \rtimes C_4)/N, $$ where $N$ is a certain normal subgroup of order two. In this case you need to construct an epimorphism $C_4 \rtimes C_4 \to G$ with kernel $N$. Now a morphism $C_4 \rtimes C_4 \to G$ is essentially the same as a pair of elements $g,h \in G$ with $g^4=h^4=1$ and $g^h = g^{-1}$. It is epimorphic if and only if $g$ and $h$ generate $G$. So you see we are doing exactly the same as before.

edited Dec 01 '17 at 15:14

answered Dec 01 '17 at 14:44

Dune

7,651

My definition is $Q_8:=\langle i,j|i^2=j^2=-1, ij=-ji\rangle$. If I take any $h\in G-\langle g\rangle$ as my $j$ and $g$ as my $i$ then $g^2=h^2=-1$ is trivial. But how can I conclude $gh=-hg$? – Buh Dec 01 '17 at 15:02
@Buh: Since $\langle h \rangle$ is normal in $G$ and $h$ is of order four, we have $h^g \in { h, h^{-1} }$. If $h^g = h$ then $G$ would be abelian, so it must be $h^g = h^{-1}$. Btw: Your definition is a bit strange. What is $-1$ or $-ji$? – Dune Dec 01 '17 at 15:11
Okay, thanks! PS: With $-1$ I meant $-1=-\mathrm{1}_2\in\text{GL}(2,\mathbb C)$. – Buh Dec 01 '17 at 15:21

Classifying groups of order 8 without using conjugation

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