# How to construct group table without the hint of subgroup $H =\{(1), (12), (34), (12)(34)\}$ is a subgroup in $S_4$

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: $$\begin{array}{|c|c|c|c|} \hline * & e& u & v & w\\ \hline e & e& u & v & w\\ \hline u & u& v & w & e\\ \hline v & v& w & e & u\\ \hline w & w& e & u & v\\ \hline \end{array}$$

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

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 $$G$$

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

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

• Your answer uses the provided fact that $v^2=e$, or $u.u^3= v^2= e.e=e$. Also, $u.v= u^3$. May 2 at 17:08
• 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. May 2 at 17:11
• 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. May 2 at 17:13
• 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. May 2 at 17:17