# Thread: Group Theory(1):Center of groups and other problems

1. ## Group Theory(1):Center of groups and other problems

Hello MHF,

Now I finished reading Artin, Chapter 2 and was attempting exercises. Few problems look really tricky. I will be grateful if anybody could help me out with these:

1) Prove that if a group contains exactly one element of order 2, then that element is in the center of the group.
2) Prove that a group whose order is a power of prime p,contains an element of order p.

Something that looked like a promising direction:If $g_i$ denotes a general element in G, I proceeded like this:

Lets say there does not exist an element of order prime. Consider the set $S = \{g_1 ^p, g_2 ^p,.....g_{p^k - 1}^p\}$ where I have not considered the identity. Then it forces S to be a permutation of G \ {e}. But I could not proceed further...
3) What is the center of the group of all unit upper triangular matrices?
Clearly it has the set of all scalar matrices(matrices of form cI) as its subgroup... But do you know any other matrix that commutes with all unit upper triangular matrices?

I would appreciate hints compared to solutions.

Thanks,
Srikanth

2. when you put your question in "Quote", we can't quote you anymore.

Question 1: suppose $a$ is the unique element of order 2 and $b$ is any element of the group. then $bab^{-1}$ has order 2 and thus $bab^{-1}=a. \ \ \Box$

Question 2: proof by induction on $|G|.$ if $|G|=p,$ then every non-identity element of G has order p and we're done. let $|G|=p^n,$ and suppose that the claim is true for any group of order

$p^m, \ 1 \leq m < n.$ if G is cyclic, then there's nothing to prove. if it's not cyclic, then choose $1 \neq x \in G.$ now apply induction on $. \ \ \Box$

Question 3: the center has only one element, i.e. the identity matrix. suppose G is the group of all $n \times n$ unit upper triangular matrices over a field or a unitary commutative ring. as usual

let $e_{ij}$ be the matrix with 1 in its (i,j)-entry and 0 elsewhere. let $g \in Z(G).$ we have $g=I + \sum_{1 \leq i < j \leq n} a_{ij}e_{ij},$ for some scalars $a_{ij}.$ now let $1 \leq k < \ell \leq n.$ then $I + e_{k \ell} \in G$ and so we must

have $(I+e_{k \ell})g=g(I+e_{k \ell}),$ which gives us: $\sum_{1 \leq i < j \leq n} a_{ij}e_{k \ell}e_{ij}=\sum_{1 \leq i < j \leq n} a_{ij}e_{ij}e_{k \ell}.$ using the fact that $e_{rs}e_{tu}=\delta_{st}e_{ru},$ we'll have: $\sum_{\ell < j \leq n} a_{\ell j}e_{kj}=\sum_{1 \leq i < k} a_{ik}e_{i \ell},$ and this gives us:

$a_{ik}=a_{\ell j} = 0, \ \ \forall 1 \leq i < k < \ell < j \leq n.$ therefore $g=I,$ and thus: $Z(G)=\{I \}. \ \ \Box$

3. Originally Posted by NonCommAlg
when you put your question in "Quote", we can't quote you anymore.
Oops! I will keep that in mind.

Originally Posted by NonCommAlg
Question 1: suppose $a$ is the unique element of order 2 and $b$ is any element of the group. then $bab^{-1}$ has order 2 and thus $bab^{-1}=a. \ \ \Box$
Thanks. Looks like we can replace 2 by any prime.

Originally Posted by NonCommAlg
Question 2: proof by induction on $|G|.$ if $|G|=p,$ then every non-identity element of G has order p and we're done. let $|G|=p^n,$ and suppose that the claim is true for any group of order

$p^m, \ 1 \leq m < n.$ if G is cyclic, then there's nothing to prove. if it's not cyclic, then choose $1 \neq x \in G.$ now apply induction on $. \ \ \Box$
This technique is nice. Thank you.I never thought of using induction.

Originally Posted by NonCommAlg
Question 3: the center has only one element, i.e. the identity matrix. suppose G is the group of all $n \times n$ unit upper triangular matrices over a field or a unitary commutative ring. as usual

let $e_{ij}$ be the element of G with 1 in its (i,j)-entry and 0 elsewhere. let $g \in Z(G).$ we have $g=I + \sum_{1 \leq i < j \leq n} a_{ij}e_{ij},$ for some scalars $a_{ij}.$ now let $1 \leq k < \ell \leq n.$ then $I + e_{k \ell} \in G$ and thus

we must have $(I+e_{k \ell})g=g(I+e_{k \ell}),$ which gives us: $\sum_{1 \leq i < j \leq n} a_{ij}e_{k \ell}e_{ij}=\sum_{1 \leq i < j \leq n} a_{ij}e_{ij}e_{k \ell}.$ using the fact that $e_{rs}e_{tu}=\delta_{st}e_{ru},$ we'll have: $\sum_{\ell < j \leq n} a_{\ell j}e_{kj}=\sum_{1 \leq i < k} a_{ik}e_{i \ell},$ and this gives us:

$a_{ik}=a_{\ell j} = 0, \ \ \forall 1 \leq i < k < \ell < j \leq n.$ therefore $g=I,$ and thus: $Z(G)=\{I \}. \ \ \Box$
I got confused with the definition of Z(G) and thats why I thought cI will work. Clearly for $c \neq 1$, cI does not belong to that group. Thanks again