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Given that $G=\{e, u,v,w\}$ is a group of order $4$ with $u^2=v, v^2=e$. Construct its multiplication table.

Does such a group exist?

The table has seven unfilled entries, that have no way to be filled with the given data.

\begin{array}{|c|c|c|c|} \hline * & e& u & v & w\\ \hline e & e& u & v & w\\ \hline u & u& v & 1 & 2\\ \hline v & v& 3 & e & 4\\ \hline w & w& 5 & 6 & 7\\ \hline \end{array}

The hint given is:

Show that $H=\{(1), (12), (34), (12)(34)\}$ is a subgroup in $S_4$, and that $\theta^2=1$ for all $\theta$ in $H$. In particular, it shows that the earlier group is associative.

But, this hint confuses as could not derive directly from given data the 7 elements.

Also, why is a subgroup used to form a group? Though it is true that being a subgroup $H$ has identity element=$(1)$, and each element has inverse, and being a subgroup has associative property wrt the composition operation. Though, composition operation is a function, hence associativity is implied.

The group table for $H$ is: \begin{array}{|c|c|c|c|} \hline * & (1)& (12) & (34) & (12)(34)\\ \hline (1) & 1& u & v & (12)(34)\\ \hline (12) & (12)& (1) & (12)(34) & (34)\\ \hline (34) & (34)& (12)(34) & (1) & (12)\\ \hline (12)(34) & (12)(34)& (34) & (12) & (1)\\ \hline \end{array}

Edit

The table has only choice available for element 3 as : $w$, hence 5=$e$. Also, 4= $u$.

The new incomplete table is: \begin{array}{|c|c|c|c|} \hline * & e& u & v & w\\ \hline e & e& u & v & w\\ \hline u & u& v & 1 & 2\\ \hline v & v& w & e & u\\ \hline w & w& e & 6 & 7\\ \hline \end{array}

Also, either $uw=e$, or $uv=e$.

If $uw=e$, then $ww= v$. So, $uv=e$ is not possible as the given column already has identity.

So, table is: \begin{array}{|c|c|c|c|} \hline * & e& u & v & w\\ \hline e & e& u & v & w\\ \hline u & u& v & w & e\\ \hline v & v& w & e & u\\ \hline w & w& e & u & v\\ \hline \end{array}

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  • $\begingroup$ @Shaun it is from book. $\endgroup$
    – jiten
    May 2 at 16:19
  • $\begingroup$ Request what is wrong in question, have tried to attempt. $\endgroup$
    – jiten
    May 2 at 16:19
  • $\begingroup$ Which book is it from? $\endgroup$
    – Shaun
    May 2 at 16:29
  • $\begingroup$ Section 2.3 Q.10 , Abstract algebra fifth edition by: Abraham Hillman, et. al. Hint reference is to Q.12 in sec 2.5. $\endgroup$
    – jiten
    May 2 at 16:35

2 Answers 2

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So, by definition of the identity $eg=g$ for any $g$ in the group. Hence $ee=e$

Next, because of the existence of inverses, if $gh=gk$ for some group elements $g,h,k$, we must have $h=k$. In particular, this implies that each element can appear at most once in each row and column.

So $uv=v$ or $uv = w$. But multiplying the first of these on the right by $v$ implies that $u=e$, which can't be. Hence $uv=w$ is the entry 1 in the second row. Then $uw=e$ is the entry 2, as that is the only entry not to appear in that row. Proceeding as such, you can uniquely fill out the table.

I don't see how the hint is helpful to you though, unless the question meant to say $u^2=e$, or $H$ were a difference subgroup

As for why considering a subgroup is helpful: a subgroup is a group in its own right. So if you could show that there was a subgroup $H$ of $S_4$ that satisfied the desired properties of $G$, that, you could take $G=H$ as the desired group. The subgroup $H$ generated by the 4-cycle $(1234)$ is an example of a group with the same multiplication table as $G$

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By Lagrange's Theorem, $\lvert u\rvert\mid \lvert G\rvert$. Thus the order of $u$, being nontrivial, is either two or four; but $u^2=v\neq e$, so $\lvert u\rvert =4$.

Consider the powers of $u$. There are four distinct powers:

$$e, u, u^2=v, u^3.$$

Thus $w=u^3$, since there are four elements of $G$.

Thus the table is

$$\begin{array}{|c|c|c|c|c|} \hline \times & e & u & v & w\\ \hline e & e & u & v & w\\ \hline u & u & v & w & e\\ \hline v & v & w & e & u\\ \hline w & w & e & u & v \\ \hline \end{array}.$$

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  • $\begingroup$ Your answer uses the provided fact that $v^2=e$, or $u.u^3= v^2= e.e=e$. Also, $u.v= u^3$. $\endgroup$
    – jiten
    May 2 at 17:08
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    $\begingroup$ Not quite, @jiten. It's more along the lines of $$u^4=u^{2\times 2}=(u^2)^2=v^2=e$$ first, but, yes, $u\times u^3=u^4$ as well. $\endgroup$
    – Shaun
    May 2 at 17:11
  • $\begingroup$ Please suggest a similar question (not considering necessarily the subgroup part, which is an analogy to just show to beginners that such group exists) for order 5,6. $\endgroup$
    – jiten
    May 2 at 17:13
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    $\begingroup$ You could try finding a generator for each of the cyclic groups of order five and six. If you want something more specific, then I suggest you ask a new, separate question, @jiten. $\endgroup$
    – Shaun
    May 2 at 17:17

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