Let $a$ and $b$ elements of order $2$ in a group $G$. Suppose $\mathrm{ord} (a,b) = 4$. Show that the subgroup generated by $a$ and $b$ is $D_4$.

Asked Jul 02 '22 at 18:13

Active Jul 02 '22 at 18:29

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Let $a$ and $b$ elements of order $2$ in a group $G$. Suppose $\mathrm{ord} (a,b) = 4$. Show that the subgroup generated by $a$ and $b$ is $D_4$.

Problem is from Bhattacharya's book.

I have a question about this question since I think there is a mistake in the order of elements. It seems to me $\mathrm{ord}(a)=4$ and $\mathrm{ord}(b)=2$ otherwise how can we conclude that this subgroup generates $D_4$.

We know

$$D_4=\langle a,b:a^4=b^2=e, ab=ba^3 \rangle.$$

If the order of $a$ is $2$ then there must be a contradiction? Any other explanation? Thanks in advance!

edited Jul 02 '22 at 18:29

Shaun

44,997

asked Jul 02 '22 at 18:13

beingmathematician

Have a look at Theorem $2.1$ in K. Conrad's notes: Let $G$ be a finite nonabelian group generated by two elements of order $2$. Then $G$ is isomorphic to a dihedral group. – Dietrich Burde Jul 02 '22 at 18:16
How could it be possible? – beingmathematician Jul 02 '22 at 18:17
2

The product of two elements of order $2$ need not have order $2$. – Dietrich Burde Jul 02 '22 at 18:33

Let $a$ and $b$ elements of order $2$ in a group $G$. Suppose $\mathrm{ord} (a,b) = 4$. Show that the subgroup generated by $a$ and $b$ is $D_4$.

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