subgroups!

[Steven G]

I can show that this is a trick question Lets call \(\displaystyle P\) the superstition that \(\displaystyle a(bc)=a\).
So we accept both \(\displaystyle a^2=b~\&~a^3=c\) as true in the following two Cayley tables:
Table I \(\displaystyle \begin{array}{*{20}{c}}
\circ &|&e&a&b&c&d \\
\hline
e&|&e&a&b&c&d \\
a&|&a&b&c&d&e \\
b&|&b&c&d&e&a \\
c&|&c&d&e&a&b \\
d&|&d&e&a&b&c \end{array}\) TABLE II \(\displaystyle \begin{array}{*{20}{c}}
\circ &|&e&a&b&c \\ \hline
e&|&e&a&b&c \\
a&|&a&b&c&e \\
b&|&b&c&e&a \\
c&|&c&e&a&b \end{array}\)
Both of the tables conform to the accepted two conditions, but only table I conforms to the other condition \(\displaystyle P:~a(bc)=a\).
Note that both tables represent a group generated by \(\displaystyle a\)
In order to accomplish that we had to introduce an element \(\displaystyle d\). Now the group is order five and each non-rival element is order five.

You really should argue that there are no larger groups whose order is more than 5 and less than 10. I do realize that this will be trivial but I fell necessary. What do you think?

 

[bZNyQ7C2]

I can show that this is a trick question Lets call \(\displaystyle P\) the superstition that \(\displaystyle a(bc)=a\).
So we accept both \(\displaystyle a^2=b~\&~a^3=c\) as true in the following two Cayley tables:
Table I \(\displaystyle \begin{array}{*{20}{c}}
\circ &|&e&a&b&c&d \\
\hline
e&|&e&a&b&c&d \\
a&|&a&b&c&d&e \\
b&|&b&c&d&e&a \\
c&|&c&d&e&a&b \\
d&|&d&e&a&b&c \end{array}\) TABLE II \(\displaystyle \begin{array}{*{20}{c}}
\circ &|&e&a&b&c \\ \hline
e&|&e&a&b&c \\
a&|&a&b&c&e \\
b&|&b&c&e&a \\
c&|&c&e&a&b \end{array}\)
Both of the tables conform to the accepted two conditions, but only table I conforms to the other condition \(\displaystyle P:~a(bc)=a\).
Note that both tables represent a group generated by \(\displaystyle a\)
In order to accomplish that we had to introduce an element \(\displaystyle d\). Now the group is order five and each non-rival element is order five.

Yes, table 1 was my solution. Once I worked out that this group is cyclical, and then a^5 is the identity (which requires what you call P) I could see that a^4 must be an element in the group, so I called it d. BTW, thanks for showing me how to make a table in LATEX.

@Jomo ... that's interesting. I interpreted the question to mean "find the smallest group that meets these conditions." I'm guessing that larger groups meeting the same conditions do exist, but that's just a guess. I would like to know one way or the other.

 

[pka]

BTW, thanks for showing me how to make a table in LATEX.

There is a wonderful piece of software MathType.
I use it for all complicated constructions.

 

[bZNyQ7C2]

There is a wonderful piece of software MathType.
I use it for all complicated constructions.

Really? I have MathType, I use it a lot. I've never been able to make a table with it though. Is there a function I'm missing?

 

[Sonal7]

I am sorry, I was out and about and didnt work on the problems any further. I went with the definition that the order of the element which gives you the identity is the order the group. The answer book gives a 2 line answer, suggesting this approach which made me uncomfortable. I do feel happier working on the Cayley table as it makes it clear to see that the order the group. I cant work on this at the moment.

 

[pka]

Really? I have MathType, I use it a lot. I've never been able to make a table with it though. Is there a function I'm missing?

One the menu bar at the top left-hand, in the second row, option second from the right is the drop-down matrix menu..
In the last row there is an option to create a matrix of any size. If you play around with the different options you will find it is really easy.
However, the vertical dividing line does not work with most compilers.

 

[Steven G]

I am sorry, I was out and about and didnt work on the problems any further. I went with the definition that the order of the element which gives you the identity is the order the group. The answer book gives a 2 line answer, suggesting this approach which made me uncomfortable. I do feel happier working on the Cayley table as it makes it clear to see that the order the group. I cant work on this at the moment.

Yes, most all beginning students would probably use the Cayley table. Can we please see the author's solution?

 

[Sonal7]

I did mean to write the power of an element that gives you the identity is the order of the element and the order of the group. The answer to the questions is: