# Explain why the example in Chapter 4 works and why their answer in Chapter 5 is invalid.

This question is from A Course in Group Theory by John Humphreys.

I have provided the components from the textbook that are necessary for the problem. I just want to know if my explanation is correct?

Question: This is a discussion about problem 5 in Exercise 5. The answer in the back of the book states that the same example in a Chapter 4 problem works for this problem, yet they change A to $\{1,a\}$ instead of $\{1,ba\}$. Explain why the example in Chapter 4 works and why their answer in Chapter 5 is invalid.

$\textbf{Chapter 5 question:}$ Give an example of a group G with subgroups A and B such that AB is not a subgroup of G.

$\textbf{Chapter 5 solution:}$ An example is provided by the same group as in the solution to exercise 2 of chapter 4. We take G to be the dihedral group $D_3$, A to be $\{1,a\}$ and B to be $\{1,b\}$ so that AB is the set $\{1,a,b,ab\}$.

$\textbf{Chapter 4 question:}$ Give an example of a group G with subgroups H and K such that $H \cup K$ is not a subgroup of G.

$\textbf{Chapter 4 solution:}$ In the dihedral group $D_3$, the set $\{a,b\}$ is a subgroup, as is $\{1,ba\}$, but the set $\{1,b,ba\}$ is not closed (here the elements a and b are generators of the dihedral group as defined in the presentation on p.33).

$\textbf{Presentation on p.33:}$ $D_3=\langle a,b:b^2=1=a^3, ab=ba^{-1} \rangle$

The reason why the answer in Chapter 5 is invalid is because their choose of the subgroup A is incorrect. If you multiply the elements a and a, you will obtain $a^2$ which is not a part of the subgroup. Hence A is not truely a subgroup because it doesn't have closure under the identity property.

• I'm sorry - I can't discern from this the questions being asked in Chapter 4 and Chapter 5 for which the solutions are given. Without the questions I cannot possibly comment on whether the solutions given are valid or not. – Mark Bennet Oct 1 '14 at 15:22
• @MarkBennet I have included the questions corresponding the solutions that were previously posted – Username Unknown Oct 1 '14 at 15:31
• $\{a,b\}$ is not a subgroup of $D_3$. It doesn't even contain the identity element $1$. – Lee Mosher Oct 1 '14 at 15:51
• Ah thats true. I didn't see that. However is that enough for the discussion that I am suppose to provide – Username Unknown Oct 1 '14 at 15:55
• Is "$\{a,b\}$ is a subgroup" a typo (yours or the book's) for "$\{1,b\}$ is a subgroup"? – bof Oct 4 '14 at 1:49

I downloaded the ebook and took a look, You are saying that solution for exercise 2 of Ch 4, {$a,b$} is a subgroup, no he is taking $H$ ={$1,b$} and $K$ ={$1,ba$} to be two subgroups. So obviously their union that is $H \cup K$= {$1,b,ba$} is not a subgroups because it does not have closure i.e. $b.ba=a \notin$$H \cup K$. Correct it.
Now He is s aying that examole is provided by same exercise $2$ of chapter $4$, and yes it is, but he did a typo here. What he was saying is take $A=${$1,b$} and $B=${$1,ba$} same as in ch $4$ ex $2$, then $AB$={$1,a,b,ba$} which is not a subgroup as $ab$ is not in $AB$ so closure is missing.